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The midpoint of a line segment divides the segment into two congruent segments. 5. Given. 6. Perpendicular lines intersect to form right angles. 7. If two angles are right angles, they are congruent. 8. ASA (steps 2, 4, 7). 1. Marty said that if two triangles are congruent and one of the triangles is a right triangle, then the other triangle must be a right triangle. Do you agree with Marty? Explain why or why not. 2. In ABC and DEF, AB > DE, and /B > /E. C F a. Dora said that if, then it follows that ABC DEF. Do you agree with Dora? Justify your answer. is congruent to DF AC A B D E b. Using only the SAS and ASA postulates, if ABC is not congruent to DEF, what sides and what angles cannot be congruent? 14365C04.pgs 7/12/07 3:05 PM Page 164 164 Congruence of Line Segments, Angles, and Triangles Developing Skills In 3–5, tell whether the triangles in each pair can be proved congruent by ASA, using only the marked congruent parts in establishing the congruence. Give the reason for your answer. 3. C F 4. C 5 In 6–8, in each case name the pair of corresponding sides or the pair of corresponding angles that would have to be proved congruent (in addition to those pairs marked congruent) in order to prove that the triangles are congruent by ASA. 6. D C 7. D B 8. D A B Applying Skills E A C A C B 9. Given: E C, EDA CDB,. EC and D is the midpoint of Prove: DAE DBC A B 10. Given: h DB h BD bisects ADC and bisects ABC. Prove: ABD CBD D A C B and h AD bisects 12. Given: DBC GFD and AE E D C 11. Given: AD'BC BAC. Prove: ADC ADB A C D B at D. bisects Prove: DFE DBA FB G F E D A B C 14365C04.pgs 7/12/07 3:05 PM Page 165 4-7 PROVING TRIANGLES CONGRUENT USING SIDE, S |
IDE, SIDE Proving Triangles Congruent Using Side, Side, Side 165 Cut three straws to any lengths and put their ends together to form a triangle. Now cut a second set of straws to the same lengths and try to form a different triangle. Repeated experiments lead to the conclusion that it cannot be done. As shown in ABC and DEF, when all three pairs of corresponding sides of a triangle are congruent, the triangles must be congruent. The truth of this statement of side-side-side or SSS triangle congruence is assumed without proof. C F A B D E Postulate 4.16 Two triangles are congruent if the three sides of one triangle are congruent, respectively, to the three sides of the other. (SSS) Thus, in ABC and DEF above, AB > DE, AC > DF, and BC > EF. It follows that ABC DEF. The postulate used here is abbreviated SSS. EXAMPLE 1 Given: Isosceles JKL with midpoint of. JL Prove: JKM LKM JK > KL and M the K Proof Prove the triangles congruent by using SSS. J M L Statements S 1. JK > KL 2. M is the midpoint of. JL S 3. JM > LM S 4. KM > KM 5. JKM LKM Reasons 1. Given. 2. Given. 3. Definition of midpoint. 4. Reflexive property of congruence. 5. SSS (steps 1, 3, 4). 14365C04.pgs 7/12/07 3:05 PM Page 166 166 Congruence of Line Segments, Angles, and Triangles Exercises Writing About Mathematics 1. Josh said that if two triangles are congruent and one of the triangles is isosceles, then the other must be isosceles. Do you agree with Josh? Explain why or why not. 2. Alvan said that if two triangles are not congruent, then at least one of the three sides of one triangle is not congruent to the corresponding side of the other triangle. Do you agree with Alvan? Justify your answer. Developing Skills In 3–5, pairs of line segments marked with the same number of strokes are congruent. A line segment marked with “” is congruent to itself by the reflexive |
property of congruence. In each case, is the given information sufficient to prove congruent triangles? 3. C A B D 4. C D F E B A 5. W Z R S T In 6–8, two sides are marked to indicate that they are congruent. Name the pair of corresponding sides that would have to be proved congruent in order to prove the triangles congruent by SSS. 6. B 7. D C 8 In 9–14, pairs of line segments marked with the same number of strokes are congruent. Pairs of angles marked with the same number of arcs are congruent. A line segment or an angle marked with “” is congruent to itself by the reflexive property of congruence. In each case, is the given information sufficient to prove congruent triangles? If so, write the abbreviation for the postulate that proves the triangles congruent. 9. B C D A 10. P S 11. D C T Q R A B 14365C04.pgs 8/2/07 5:40 PM Page 167 12. C 13. A M B A D B Chapter Summary 167 C 14. S R P Q 15. If two sides and the angle opposite one of those sides in a triangle are congruent to the corresponding sides and angle of another triangle, are the two triangles congruent? Justify your answer or draw a counterexample proving that they are not. (Hint: Could one triangle be an acute triangle and the other an obtuse triangle?) Applying Skills 16. Given: AEB bisects, CED AC'CED, and BD'CED. Prove: EAC EBD 17. Given: ABC is equilateral, D is the midpoint of AB. Prove: ACD BCD 18. Given: Triangle PQR with S on PQ and RS'PQ ; PSR is not congruent to QSR. Prove: PS QS 19. Gina is drawing a pattern for a kite. She wants it to consist of two congruent triangles that share a common side. She draws an angle with its vertex at A and marks two points, B and C, one on each of the rays of the angle. Each point, B and C, is 15 inches from the vertex of the angle. Then she draws the bisector of BAC, marks a point D |
on the angle bisector and. Prove that the triangles that she drew are congruent. draws and BD CD CHAPTER SUMMARY Definitions to Know • Adjacent angles are two angles in the same plane that have a common vertex and a common side but do not have any interior points in common. • Complementary angles are two angles the sum of whose degree measures is 90. • Supplementary angles are two angles the sum of whose degree measures is 180. • A linear pair of angles are two adjacent angles whose sum is a straight angle. • Vertical angles are two angles in which the sides of one angle are opposite rays to the sides of the second angle. 14365C04.pgs 8/2/07 5:41 PM Page 168 168 Congruence of Line Segments, Angles, and Triangles • Two polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent. Corresponding parts of congruent polygons are congruent. Corresponding parts of congruent polygons are equal in measure. Postulates 4.1 A line segment can be extended to any length in either direction. 4.2 Through two given points, one and only one line can be drawn. (Two points determine a line.) 4.3 Two lines cannot intersect in more than one point. 4.4 One and only one circle can be drawn with any given point as center and the length of any given line segment as a radius. 4.5 At a given point on a given line, one and only one perpendicular can be drawn to the line. 4.6 From a given point not on a given line, one and only one perpendicular can be drawn to the line. 4.7 For any two distinct points, there is only one positive real number that is the length of the line segment joining the two points. (Distance Postulate) 4.8 The shortest distance between two points is the length of the line segment joining these two points. 4.9 A line segment has one and only one midpoint. 4.10 An angle has one and only one bisector. 4.11 Any geometric figure is congruent to itself. (Reflexive Property) 4.12 A congruence may be expressed in either order. (Symmetric Property) 4.13 Two geometric figures congruent to the same geometric figure are congru- ent |
to each other. (Transitive Property) 4.14 Two triangles are congruent if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of the other. (SAS) 4.15 Two triangles are congruent if two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the other. (ASA) 4.16 Two triangles are congruent if the three sides of one triangle are congru- ent, respectively, to the three sides of the other. (SSS) 4.1 If two angles are right angles, then they are congruent. 4.2 If two angles are straight angles, then they are congruent. 4.3 If two angles are complements of the same angle, then they are congruent. 4.4 If two angles are congruent, then their complements are congruent. 4.5 If two angles are supplements of the same angle, then they are congruent. 4.6 If two angles are congruent, then their supplements are congruent. 4.7 If two angles form a linear pair, then they are supplementary. 4.8 If two lines intersect to form congruent adjacent angles, then they are perpendicular. 4.9 If two lines intersect, then the vertical angles are congruent. Theorems 14365C04.pgs 7/12/07 3:05 PM Page 169 Review Exercises 169 VOCABULARY 4-1 Distance postulate 4-3 Adjacent angles • Complementary angles • Complement • Supplementary angles • Supplement • Linear pair of angles • Vertical angles 4-4 One-to-one correspondence • Corresponding angles • Corresponding sides • Congruent polygons 4-5 SAS triangle congruence 4-6 ASA triangle congruence 4-7 SSS triangle congruence REVIEW EXERCISES 1. The degree measure of an angle is 15 more than twice the measure of its complement. Find the measure of the angle and its complement. 2. Two angles, LMP and PMN, are a linear pair of angles. If the degree measure of LMP is 12 less than three times that of PMN, find the measure of each angle. 3. Triangle JKL is congruent to triangle PQR and mK 3a 18 and mQ 5 |
a 12. Find the measure of K and of Q. 4. If and ABC DBE tulate that justifies your answer. intersect at F, what is true about B and F? State a pos- 5. If g LM'MN and g KM'MN, what is true about g LM and g KM? State a postulate that justifies your answer. 6. Point R is not on LMN. Is LM MN less than, equal to, or greater than LR RN? State a postulate that justifies your answer. 7. If h BD and h BE are bisectors of ABC, does E lie on g BD? State a postulate that justifies your answer. 8. The midpoint of AB is M. If g MN and g PM are bisectors of AB, does P lie g MN on? Justify your answer. 9. The midpoint of AB is M. If g MN and g PM are perpendicular to AB, does P lie on? Justify your answer. g MN 14365C04.pgs 8/2/07 5:41 PM Page 170 170 Congruence of Line Segments, Angles, and Triangles 10. Given: mA 50, mB 45, AB = 10 cm, mD 50, mE 45, and DE 10 cm. Prove: ABC DEF E 11. Given: bisects GEH mD mF. DEF and Prove: GFE HDE D H C A B D F G 12. Given: AB > DE not congruent to DEF. BC > EF,, ABC is E F C F Prove: B is not congruent to E. Exploration A B D E 1. If three angles of one triangle are congruent to the corresponding angles of another triangle, the triangles may or may not be congruent. Draw diagrams to show that this is true. 2. STUVWXYZ is a cube. Write a paragraph proof that would convince someone that STX, UTX, and STU are all congruent to one another. Z V W S X T Y U CUMULATIVE REVIEW CHAPTERS 1–4 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following is an illustration of the associative property of addition? (1) 3(4 7) 3(7 4) (2) 3(4 7) 3 |
(4) 3(7) (3) 3 (4 7) 3 (7 4) (4) 3 (4 7) (3 4) 7 14365C04.pgs 7/12/07 3:05 PM Page 171 Cumulative Review 171 2. If the sum of the measures of two angles is 90, the angles are (1) supplementary. (2) complementary. (3) a linear pair. (4) adjacent angles. 3. If AB BC AC, which of the following may be false?. AC (1) B is the midpoint of (2) B is a point of AC. (3) B is between A and C. h BC h BA and (4) are opposite rays. 4. If b is a real number, then b has a multiplicative inverse only if (1) b 1 (2) b 0 (3) b 0 (4) b 0 5. The contrapositive of “Two angles are congruent if they have the same measures” is (1) Two angles are not congruent if they do not have the same measures. (2) If two angles have the same measures, then they are congruent. (3) If two angles are not congruent, then they do not have the same measures. (4) If two angles do not have the same measures, then they are not congruent. 6. The statement “Today is Saturday and I am going to the movies” is true. Which of the following statements is false? (1) Today is Saturday or I am not going to the movies. (2) Today is not Saturday or I am not going to the movies. (3) If today is not Saturday, then I am not going to the movies. (4) If today is not Saturday, then I am going to the movies. 7. If ABC BCD, then ABC and BCD must be (2) scalene. (3) isosceles. (4) equilateral. (1) obtuse. g ABC 8. If and g DBE (1) congruent vertical angles. (2) supplementary vertical angles. intersect at B, ABD and CBE are (3) congruent adjacent angles. (4) supplementary adjacent angles. 9. LMN and NMP form a linear pair of angles. Which of the following statements is false? (1) mLMN mNMP |
180 (2) LMN and NMP are supplementary angles. h ML h ML (3) (4) and and h MP h MN are opposite rays. are opposite rays. 10. The solution set of the equation 3(x 2) 5x is (1) {x x 3} (2) {x x 3} (3) {x x 1} (4) {x x 1} 14365C04.pgs 7/12/07 3:05 PM Page 172 172 Congruence of Line Segments, Angles, and Triangles Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Given: bisects PQ RS and R S. at M 12. Given: Quadrilateral DEFG with DE DG and EF GF. Prove: RMQ SMP Prove: DEF DGF P E R M S G D F Q Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The following statements are true: If our team does not win, we will not celebrate. We will celebrate or we will practice. We do not practice. Did our team win? Justify your answer. 14. The two angles of a linear pair of angles are congruent. If the measure of one angle is represented by 2x y and the measure of the other angle by x 4y, find the values of x and of y. 14365C04.pgs 7/12/07 3:05 PM Page 173 Cumulative Review 173 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Josie is making a pattern for quilt pieces. One pattern is a right triangle with two acute angles that are complementary. The measure of one of the acute angles is to be 12 degrees more than half the measure of the other acute angle. |
Find the measure of each angle of the triangle. 16. Triangle DEF is equilateral and equiangular. The midpoint of is L. Line segments is N, and of DE are drawn., and MN ML FD NL EF, is M, of a. Name three congruent triangles. b. Prove that the triangles named in a are congruent. c. Prove that NLM is equilateral. d. Prove that NLM is equiangular. 14365C05.pgs 7/10/07 8:41 AM Page 174 CHAPTER 5 CHAPTER TABLE OF CONTENTS 5-1 Line Segments Associated with Triangles 5-2 Using Congruent Triangles to Prove Line Segments Congruent and Angles Congruent 5-3 Isosceles and Equilateral Triangles 5-4 Using Two Pairs of Congruent Triangles 5-5 Proving Overlapping Triangles Congruent 5-6 Perpendicular Bisector of a Line Segment 5-7 Basic Constructions Chapter Summary Vocabulary Review Exercises Cumulative Review 174 CONGRUENCE BASED ON TRIANGLES The SSS postulate tells us that a triangle with sides of given lengths can have only one size and shape. Therefore, the area of the triangle is determined. We know that the area of a triangle is one-half the product of the lengths of one side and the altitude to that side. But can the area of a triangle be found using only the lengths of the sides? A formula to do this was known by mathematicians of India about 3200 B.C. In the Western world, Heron of Alexandria, who lived around 75 B.C., provided in his book Metrica a formula that we now call Heron’s formula: If A is the area of the triangle with sides of length a, b, and c, and the semiperimeter, s, is one-half the perimeter, that is, s (a b c), then 1 2 A 5 " s(s 2 a)(s 2 b)(s 2 c) In Metrica, Heron also provided a method of finding the approximate value of the square root of a number. This method, often called the divide and average method, continued to be used until calculators made the pencil and paper computation of a square root unnecessary. 14365C05.pgs 7/10/07 8:41 AM Page 175 5-1 LINE |
SEGMENTS ASSOCIATED WITH TRIANGLES Line Segments Associated with Triangles 175 Natalie is planting a small tree. Before filling in the soil around the tree, she places stakes on opposite sides of the tree at equal distances from the base of the tree. Then she fastens cords from the same point on the trunk of the tree to the stakes. The cords are not of equal length. Natalie reasons that the tree is not perpendicular to the ground and straightens the tree until the cords are of equal lengths. Natalie used her knowledge of geometry to help her plant the tree. What was the reasoning that helped Natalie to plant the tree? Geometric shapes are all around us. Frequently we use our knowledge of geometry to make decisions in our daily life. In this chapter you will write formal and informal proofs that will enable you to form the habit of looking for logical relationships before making a decision. Altitude of a Triangle DEFINITION An altitude of a triangle is a line segment drawn from any vertex of the triangle, perpendicular to and ending in the line that contains the opposite side In ABC, if K AB CD CD GH g EF, then, then is perpendicular to is the is perpendicular to altitude from vertex C to the opposite side. In EFG, if, the line that contains the side GH is the altitude from vertex G to the opposite EF side. In an obtuse triangle such as EFG above, the altitude from each of the acute angles lies outside the triangle. In right TSR, if is the altitude from vertex R to the opposite side. RS In a right triangle such as TSR above, the altitude from each vertex of an acute angle is a leg of the triangle. Every triangle has three altitudes as shown in JKL. is perpendicular to TS TS is the altitude from T to the opposite side, then and RS RS TS L J Median of a Triangle DEFINITION A median of a triangle is a line segment that joins any vertex of the triangle to the midpoint of the opposite side. 14365C05.pgs 7/10/07 8:41 AM Page 176 176 Congruence Based on Triangles angle bisector P D Q In ABC, if M is the midpoint of is the median drawn from CM. We may also draw a median from vertex A to the midpoint. Thus, every, and a median from vertex B to the midpoint of side, then AB AC AB vertex C to side of side BC triangle has three medians. Angle |
Bisector of a Triangle DEFINITION An angle bisector of a triangle is a line segment that bisects any angle of the triangle and terminates in the side opposite that angle. In PQR, if D is a point on is the angle bisector from R in PQR. We may also draw an angle bisector from the, and an angle bisector from the vertex Q to some vertex P to some point on point on. Thus, every triangle has three angle bisectors. such that PRD QRD, then RD QR PQ PR In a scalene triangle, the altitude, the median, and the angle bisector drawn from any common vertex are three distinct line segments. In ABC, from the common vertex B, three line segments are drawn: B 1. 2. 3. is the altitude from B because BD BD'AC. is the angle bisector from B BE because ABE EBC. BF is the midpoint of is the median from B because F AC. A F E D C median angle bisector altitude In some special triangles, such as an isosceles triangle and an equilateral triangle, some of these segments coincide, that is, are the same line. We will consider these examples later. EXAMPLE 1 Given: KM and is the angle bisector from K in JKL, LK > JK. K Prove: JKM LKM L M J 14365C05.pgs 7/10/07 8:41 AM Page 177 Line Segments Associated with Triangles 177 Proof Statements Reasons 1. LK > JK 2. is the angle bisector from KM K in JKL. 1. Given. 2. Given. 3. KM bisects JKL. 3. Definition of an angle bisector of a triangle. 4. JKM LKM 4. Definition of the bisector of an angle. KM > KM 5. 6. JKM LKM 5. Reflexive property of congruence. 6. SAS (steps 1, 4, 5). Exercises Writing About Mathematics 1. Explain why the three altitudes of a right triangle intersect at the vertex of the right angle. 2. Triangle ABC is a triangle with C an obtuse angle. Where do the lines containing the three altitudes of the triangle intersect? Developing Skills 3. Use a pencil, ruler, and protractor, or use geometry software, to draw ABC, an acute, |
sca- lene triangle with altitude CD, angle bisector CE, and median. CF a. Name two congruent angles that have their vertices at C. b. Name two congruent line segments. c. Name two perpendicular line segments. d. Name two right angles. 4. Use a pencil, ruler, and protractor, or use geometry software, to draw several triangles. Include acute, obtuse, and right triangles. a. Draw three altitudes for each triangle. b. Make a conjecture regarding the intersection of the lines containing the three altitudes. 5. Use a pencil, ruler, and protractor, or use geometry software, to draw several triangles. Include acute, obtuse, and right triangles. a. Draw three angle bisectors for each triangle. b. Make a conjecture regarding the intersection of these three angle bisectors. 14365C05.pgs 7/10/07 8:41 AM Page 178 178 Congruence Based on Triangles 6. Use a pencil, ruler, and protractor, or use geometry software, to draw several triangles. Include acute, obtuse, and right triangles. a. Draw three medians to each triangle. b. Make a conjecture regarding the intersection of these three medians. In 7–9, draw and label each triangle described. Complete each required proof in two-column format. 7. Given: In PQR, PR > QR, P Q, and RS Prove: PSR QSR is a median. 8. Given: In DEF, EG bisector and an altitude. is both an angle Prove: DEG FEG 9. Given: CD Prove: CD is a median of ABC but ADC is not congruent to BDC. is not an altitude of ABC. (Hint: Use an indirect proof.) Applying Skills In 10–13, complete each required proof in paragraph format. 10. In a scalene triangle, LNM, show that an altitude, NO, cannot be an angle bisector. (Hint: Use an indirect proof.) 11. A telephone pole is braced by two wires that are fastened to the pole at point C and to the. If the ground at points A and B. The base of the pole is at point D, the midpoint of pole is perpendicular to the ground, are the wires of equal length? Justify your answer. AB 12. The formula for the area of a triangle is A |
and h the length of the altitude to that side. In ABC, side and M is the midpoint of angles of equal area, AMC and BMC. AB AB 1 2bh with b the length of one side of a triangle is the altitude from vertex C to. Show that the median separates ABC into two tri- CD 13. A farmer has a triangular piece of land that he wants to separate into two sections of equal area. How can the land be divided? 5-2 USING CONGRUENT TRIANGLES TO PROVE LINE SEGMENTS CONGRUENT AND ANGLES CONGRUENT The definition of congruent triangles tells us that when two triangles are congruent, each pair of corresponding sides are congruent and each pair of corresponding angles are congruent. We use three pairs of corresponding parts, SAS, ASA, or SSS, to prove that two triangles are congruent. We can then conclude that each of the other three pairs of corresponding parts are also congruent. In this section we will prove triangles congruent in order to prove that two line segments or two angles are congruent. 14365C05.pgs 7/10/07 8:41 AM Page 179 Using Congruent Triangles to Prove Line Segments Congruent and Angles Congruent 179 EXAMPLE 1 Given:, A is a right angle, AE > DF, ABCD D is a right angle, AB > CD. Prove: EC > FB E F Proof The line segments that we want to prove congruent are corresponding sides of EAC and FDB. Therefore we will first prove that EAC FDB. Then use that corresponding parts of congruent triangles are congruent. A C B D Statements Reasons 1. A is a right angle, D is a 1. Given. right angle. 2. A D 3. AE > DF 4. AB > CD 2. If two angles are right angles, then they are congruent. (Theorem 4.1) 3. Given. 4. Given. 5. AB 1 BC > BC 1 CD 5. Addition postulate. 6. ABCD 7. AB 1 BC 5 AC BC 1 CD 5 BD AC > BD 8. 9. EAC FDB 10. EC > FB 6. Given. 7. Partition postulate. 8. Substitution postulate (steps 5, 7). 9. SAS (steps 3, 2, 8 |
). 10. Corresponding parts of congruent triangles are congruent. Exercises Writing About Mathematics 1. Triangles ABC and DEF are congruent triangles. If A and B are complementary angles, are D and E also complementary angles? Justify your answer. 2. A leg and the vertex angle of one isosceles triangle are congruent respectively to a leg and the vertex angle of another isosceles triangle. Is this sufficient information to conclude that the triangles must be congruent? Justify your answer. 14365C05.pgs 7/10/07 8:41 AM Page 180 180 Congruence Based on Triangles Developing Skills In 3–8, the figures have been marked to indicate pairs of congruent angles and pairs of congruent segments. a. In each figure, name two triangles that are congruent. b. State the reason why the triangles are congruent. c. For each pair of triangles, name three additional pairs of parts that are congruent because they are corresponding parts of congruent triangles. 3. M R P S 6. D C E A B 9. Given: CA > CB of. AB Prove: A B C 4. D A C B 7. X S R P Q Y 5. D C E A B 8. D C E F A B and D is the midpoint 10. Given: Prove: and AB > CD CAB ACD AD > CB D C A D B A B CED 11. Given: and AEB other. Prove: C D B C E D A bisect each 12. Given: KLM and NML are right angles and KL = NM. Prove: K N K L N M 14365C05.pgs 7/10/07 8:41 AM Page 181 Isosceles and Equilateral Triangles 181 13. Triangle ABC is congruent to triangle DEF, AB 3x 7, DE 5x 9, and BC 4x. Find: a. x b. AB c. BC d. EF 14. Triangle PQR is congruent to triangle LMN, mP 7a, mL 4a 15, and P and Q are complementary. Find: a. a b. mP c. mQ d. mM Applying Skills In 15 and 16, complete each required proof in paragraph format. 15. a. Prove that the median from the vertex angle of an |
isosceles triangle separates the triangle into two congruent triangles. b. Prove that the two congruent triangles in a are right triangles. 16. a. Prove that if each pair of opposite sides of a quadrilateral are congruent, then a diagonal of the quadrilateral separates it into two congruent triangles. b. Prove that a pair of opposite angles of the quadrilateral in a are congruent. In 17 and 18, complete each required proof in two-column format. 17. a. Points B and C separate into three congruent segments. P is a point not on such that PA > PD and. Draw a diagram that shows these line segments and ABCD PB > PC g AD write the information in a given statement. b. Prove that APB DPC. c. Prove that APC DPB. 18. The line segment PM is both the altitude and the median from P to LN in LNP. a. Prove that LNP is isosceles. b. Prove that PM is also the angle bisector from P in LNP. 5-3 ISOSCELES AND EQUILATERAL TRIANGLES When working with triangles, we observed that when two sides of a triangle are congruent, the median, the altitude, and the bisector of the vertex angle separate the triangle into two congruent triangles. These congruent triangles can be used to prove that the base angles of an isosceles triangle are congruent. This observation can be proved as a theorem called the Isosceles Triangle Theorem. 14365C05.pgs 7/10/07 8:41 AM Page 182 182 Congruence Based on Triangles Theorem 5.1 If two sides of a triangle are congruent, the angles opposite these sides are congruent. Given ABC with AC > BC Prove A B C Proof In order to prove this theorem, we will use the median to the base to separate the triangle into two congruent triangles. A D B Statements Reasons 1. Draw D, the midpoint of. AB 1. A line segment has one and only one midpoint. 2. is the median from CD vertex C. 2. Definition of a median of a triangle. 3. CD > CD 4. AD > DB AC > BC 5. 6. ACD BCD 7. A B 3. Reflexive property of con |
gruence. 4. Definition of a midpoint. 5. Given. 6. SSS (steps 3, 4, 5). 7. Corresponding parts of congruent triangles are congruent. A corollary is a theorem that can easily be deduced from another theorem. We can prove two other statements that are corollaries of the isosceles triangle theorem because their proofs follow directly from the proof of the theorem. Corollary 5.1a The median from the vertex angle of an isosceles triangle bisects the vertex angle. Proof: From the preceding proof that ACD BCD, we can also conclude that ACD BCD since they, too, are corresponding parts of congruent triangles. Corollary 5.1b The median from the vertex angle of an isosceles triangle is perpendicular to the base. 14365C05.pgs 7/10/07 8:41 AM Page 183 Isosceles and Equilateral Triangles 183 Proof: Again, from ACD BCD, we can say that CDA CDB because they are corresponding parts of congruent triangles. If two lines intersect to form congruent adjacent angles, then they are perpendicular. Therefore, CD'AB. Properties of an Equilateral Triangle The isosceles triangle theorem has shown that in an isosceles triangle with two congruent sides, the angles opposite these sides are congruent. We may prove another corollary to this theorem for any equilateral triangle, where three sides are congruent. Corollary 5.1c Every equilateral triangle is equiangular. Proof: If ABC is equilateral, then By the isosceles triangle theorem, since A C, and since fore, A B C. AB > BC > CA. AB > BC,, B A. There- BC > CA Given: E not on ABCD, AB > CD, and EB > EC. Prove: AE > DE C B A E DE and are corresponding sides of AE ABE and DCE, and we will prove these triangles congruent by SAS. We are given two pairs of congruent corresponding sides and must prove that the included angles are congruent. A B C D EXAMPLE 1 Proof E Statements EB > EC 1. 2. EBC ECB A B C D 3. ABCD Reasons 1. Given. 2. Isosceles triangle theorem (Or: If two sides of a |
triangle are congruent, the angles opposite these sides are congruent.). 3. Given. Continued 14365C05.pgs 7/10/07 8:41 AM Page 184 184 Congruence Based on Triangles (Continued) E Statements 4. ABE and EBC are supplementary. DCE and ECB are supplementary. 5. ABE DCE A B C D AB > CD 6. 7. ABE DCE 8. AE > DE Reasons 4. If two angles form a linear pair, they are supplementary. 5. The supplements of congruent angles are congruent. 6. Given. 7. SAS (steps 1, 5, 6). 8. Corresponding parts of congruent triangles are congruent. Exercises Writing About Mathematics 1. Joel said that the proof given in Example 1 could have been done by proving that ACE DBE. Do you agree with Joel? Justify your answer. 2. Abel said that he could prove that equiangular triangle ABC is equilateral by drawing and showing that ABD CBD. What is wrong with Abel’s reasoning? median BD Developing Skills 3. In ABC, if 4. Triangle RST is an isosceles right triangle with RS ST and R and T complementary, mB 3x 15 and mC 7x 5, find mB and mC. AB > AC angles. What is the measure of each angle of the triangle? 5. In equilateral DEF, mD 3x y, mE 2x 40, and mF 2y. Find x, y, mD, mE, and mF. DABE 6. Given: C not on CA > CB Prove: CAD CBE and D A B E C 7. Given: Quadrilateral ABCD with AD > CD AB > CB Prove: BAD BCD and A B D C 14365C05.pgs 7/10/07 8:41 AM Page 185 Isosceles and Equilateral Triangles 185 8. Given: AC > CB Prove: CDE CED and DA > EB 9. Given: AC > BC Prove: CAD CBD and DAB DBA C D E C D A B A B 10. Given: Isosceles ABC with AC > BC and F is the midpoint of, D is the midpoint of AB. BC E is the midpoint of a. Prove: ADF BEF b. Prove: DEF is |
isosceles. AC, C D E A F B In 11 and 12, complete each given proof with a partner or in a small group. 11. Given: ABC with AB AC, BG EC,, and 12. Given: E not on BE'DE CG'GF and EB. AB > CD,, ABCD is not congruent to. EC Prove: BD > CF Prove: AE is not congruent to DE. A D F B E G C Applying Skills E A B C D (Hint: Use an indirect proof.) 13. Prove the isosceles triangle theorem by drawing the bisector of the vertex angle instead of the median. 14. Prove that the line segments joining the midpoints of the sides of an equilateral triangle are congruent. › ‹ 15. C is a point not on FBDG 16. In PQR, mR mQ. Prove that PQ PR. and BC DC. Prove that FBC GDC. 14365C05.pgs 7/10/07 8:41 AM Page 186 186 Congruence Based on Triangles 5-4 USING TWO PAIRS OF CONGRUENT TRIANGLES Often the line segments or angles that we want to prove congruent are not corresponding parts of triangles that can be proved congruent by using the given information. However, it may be possible to use the given information to prove a different pair of triangles congruent. Then the congruent corresponding parts of this second pair of triangles can be used to prove the required triangles congruent. The following is an example of this method of proof. EXAMPLE 1 Given:, AEB AC > AD, and CB > DB Prove: CE > DE C A B E Proof Since CE DE and are corresponding parts of ACE and ADE, we can prove these two line segments congruent if we can prove ACE and ADE congruent. From the given, we cannot prove immediately that ACE and ADE congruent. However, we can prove that CAB DAB. Using corresponding parts of these larger congruent triangles, we can then prove that the smaller triangles are congruent. D Statements Reasons A A C D C D 1. AC > AD 2. CB > DB 3. AB > AB B E 4. CAB DAB 5. CAB DAB B 6. AE > AE E 7. CA |
E DAE 8. CE > DE 1. Given. 2. Given. 3. Reflexive property of congruence. 4. SSS (steps 1, 2, 3). 5. Corresponding parts of congruent triangles are congruent. 6. Reflexive property of congruence. 7. SAS (steps 1, 5, 6). 8. Corresponding parts of congruent triangles are congruent. 14365C05.pgs 7/10/07 8:41 AM Page 187 Using Two Pairs of Congruent Triangles 187 Exercises Writing About Mathematics 1. Can Example 1 be proved by proving that BCE BDE? Justify your answer. 2. Greg said that if it can be proved that two triangles are congruent, then it can be proved that the medians to corresponding sides of these triangles are congruent. Do you agree with Greg? Explain why or why not. Developing Skills 3. Given: ABC DEF, M is the, and N is. DE midpoint of AB the midpoint of Prove: AMC DNF B M A C F 4. Given: ABC DEF, bisects ACB, and bisects DFE. CG FH Prove: CG > FH. Given: AEC FEG and DEB intersects bisect each other, at G and AB CD at F. 6. Given: AME BMF and DE > CF Prove: E is the midpoint of. FEG Prove: AD > BC 14365C05.pgs 7/10/07 8:41 AM Page 188 188 Congruence Based on Triangles 7. Given: Prove: BC > BA h DB bisects CDA. h BD and bisects CBA. 8. Given: RP RQ and SP SQ Prove: RT'PQ R S P T Q B A C D Applying Skills 9. In quadrilateral ABCD, AB = CD, BC = DA, and M is the midpoint of through M intersects AB at E and CD at F. Prove that BMD bisects 10. Complete the following exercise with a partner or in a small group: BD EMF. A line segment at M. Line l intersects AB SA = SB, then M is the midpoint of AB and l is perpendicular to AB. at M, and P and S are any two points on l. Prove that if PA = |
PB and a. Let half the group treat the case in which P and S are on the same side of AB. b. Let half the group treat the case in which P and S are on opposite sides of AB. c. Compare and contrast the methods used to prove the cases. 5-5 PROVING OVERLAPPING TRIANGLES CONGRUENT AD > BC DB > CA, can we prove If we know that and that DBA CAB? These two triangles overlap and share a common side. To make it easier to visualize the overlapping triangles that we want to prove congruent, it may be helpful to outline each with a different color as shown in the figure. AB Or the triangles can be redrawn as separate triangles. is a side of The segment each of the triangles DBA and CAB. Therefore, to the given inforDB > CA mation,, we and prove that can add DBA CAB by SSS. AB > AB AD > BC and 14365C05.pgs 7/10/07 8:41 AM Page 189 EXAMPLE 1 Given: BE and are medians CD to the legs of isosceles ABC. Proving Overlapping Triangles Congruent 189 A A A D E DE Prove: CD > BE B BC C Proof Separate the triangles to see more clearly the triangles to be proved congruent. We know that the legs of an isosceles triangle are congruent. Therefore, AB > AC. We also know that the median is a line segment from a vertex to the midpoint of the opposite side. Therefore, D and E are midpoints of the congruent legs. The midpoint divides the line segment into two congruent segments, that is, in half, and halves of congruent segments are congruent: AE > AD AE > AD is A, and A is congruent to itself by the reflexive property of congruence. Therefore, ABE ACD by SAS and because corresponding parts of congruent triangles are congruent.. Now we have two pair of congruent sides:. The included angle between each of these pairs of congruent sides AB > AC CD > BE and EXAMPLE 2 Solution Using the results of Example 1, find the length of CD x 15. BE if BE 5x 9 and BE CD 5x 9 x 15 4x 24 x 6 BE 5x 9 5(6) 9 30 9 |
21 CD x 15 6 15 21 Answer 21 Exercises Writing About Mathematics 1. In Example 1, the medians to the legs of isosceles ABC were proved to be congruent by proving ABE ACD. Could the proof have been done by proving DBC ECB? Justify your answer. 14365C05.pgs 7/10/07 8:41 AM Page 190 190 Congruence Based on Triangles 2. In Corollary 5.1b, we proved that the median to the base of an isosceles triangle is also the altitude to the base. If the median to a leg of an isosceles triangle is also the altitude to the leg of the triangle, what other type of triangle must this triangle be? Developing Skills 3. Given: AE > FB, DA > CB,, AEFB and A and B are right angles. DF > CE Prove: DAF CBE and 4. Given: SPR > SQT, Prove: SRQ STP and R T PR > QT. Given: DA > CB CB'AB, DA'AB, and 6. Given: AC > BD Prove:, BAE CBF, ABCD BCE CDF, AE > BF and E F AB > CD Prove: DAB CBA and TR 8. Given: AD > CE Prove: ADC CEA and DB > EB 7. Given: TM > TN and N is the midpoint of RN > SM Prove:, M is the midpoint of. TS Applying Skills In 9–11, complete each required proof in paragraph format. 9. Prove that the angle bisectors of the base angles of an isosceles triangle are congruent. 10. Prove that the triangle whose vertices are the midpoints of the sides of an isosceles triangle is an isosceles triangle. 11. Prove that the median to any side of a scalene triangle is not the altitude to that side. 14365C05.pgs 7/10/07 8:41 AM Page 191 Perpendicular Bisector of a Line Segment 191 5-6 PERPENDICULAR BISECTOR OF A LINE SEGMENT Perpendicular lines were defined as lines that intersect to form right angles. We also proved that if two lines intersect to form congruent adjacent angles, then they are perpendicular. (Theorem 4.8) The bisector of |
a line segment was defined as any line or subset of a line that intersects a line segment at its midpoint In the diagrams, g, PM g, NM QM, and h MR are all bisectors of AB since they each intersect AB at its midpoint, M. Only is both perpendicular to AB and g NM g NM is the perpendicular bisector of. AB the bisector of AB. DEFINITION The perpendicular bisector of a line segment is any line or subset of a line that is perpendicular to the line segment at its midpoint. In Section 3 of this chapter, we proved as a corollary to the isosceles triangle theorem that the median from the vertex angle of an isosceles triangle is perpendicular to the base. In the diagram below, since g CM CM'AB. Therefore, ABC, CM is the median to the base of isosceles is the perpendicular bisector of. AB (1) M is the midpoint of AB : AM = MB. C M is equidistant, or is at an equal distance, from the endpoints of AB. (2) AB is the base of isosceles ABC: AC = BC. C is equidistant from the endpoints of AB. These two points, M and C, determine the perpendicular bisector of. This suggests the following theorem. AB A M B 14365C05.pgs 7/10/07 8:41 AM Page 192 192 Congruence Based on Triangles Theorem 5.2 If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment. Given and points P and T such that PA = PB and AB TA = TB. Prove g PT is the perpendicular bisector of. AB A Strategy Let g PT AB intersect at M. Prove APT BPT by SSS. Then, using the congruent corresponding angles, prove APM BPM by SAS. Consequently, is a bisector. Also, AMP BMP. AM > MB Since two lines that intersect to form congruent adjacent g PT, so angles are perpendicular, the perpendicular bisector of g AB'PT AB.. Therefore, g PT is B P T A B P M T The details of the proof of Theorem 5.2 will be left to the student. (See exercise 7.) Theorem 5.3a If a point is equidistant from the end |
points of a line segment, then it is on the perpendicular bisector of the line segment. Given Point P such that PA = PB. Prove P lies on the perpendicular bisector of. AB Proof Choose any other point that is equidistant from AB. Then AB by Theorem 5.2. (If two points, for example, M, the midg is the perpendicular PM the endpoints of point of AB bisector of are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment.) P lies on g PM. The converse of this theorem is also true. P P B A A M B 14365C05.pgs 7/10/07 8:41 AM Page 193 Theorem 5.3b If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. Perpendicular Bisector of a Line Segment 193 Given Point P on the perpendicular bisector of. AB Prove PA = PB AB Proof Let M be the midpoint of PM > PM AM > BM. Then. Perpendicular lines intersect to and form right angles, so PMA PMB. By SAS, PMA PMB. Since corresponding parts of congruent triangles are congruent, PA PB. PA > PB and C P M D B A Theorems 5.3a and 5.3b can be written as a biconditional. Theorem 5.3 A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the endpoints of the line segment. Methods of Proving Lines or Line Segments Perpendicular To prove that two intersecting lines or line segments are perpendicular, prove that one of the following statements is true: 1. The two lines form right angles at their point of intersection. 2. The two lines form congruent adjacent angles at their point of intersection. 3. Each of two points on one line is equidistant from the endpoints of a seg- ment of the other. Intersection of the Perpendicular Bisectors of the Sides of a Triangle When we draw the three perpendicular bisectors of the sides of a triangle, it appears that the three lines are concurrent, that is, they intersect in one point. Theorems 5.3a and 5.3b allow us to prove the following perpendicular bisector conc |
urrence theorem. C N P L A M B 14365C05.pgs 7/10/07 8:41 AM Page 194 194 Congruence Based on Triangles Theorem 5.4 The perpendicular bisectors of the sides of a triangle are concurrent. Given g MQ, the perpendicular bisector of AB C C Proof EXAMPLE 1 g NR, the perpendicular bisector of AC g LS, the perpendicular bisector of BC Prove g, MQ g NR, and g LS intersect in P MQ A (1) We can assume from the diagram that and g NR intersect. Let us call the point of intersection P. (2) By theorem 5.3b, since P is a point on g MQ, the perpendicular bisector of AB, P is equidistant from A and B. (3) Similarly, since P is a point on g NR, the perpendicular bisector of AC, P is equidistant from A and C. (4) In other words, P is equidistant from A, B, and C. By theorem 5.3a, since P, P is on the perpendicular bisector BC is equidistant from the endpoints of of BC. (5) Therefore, g MQ, g NR, and g LS, the three perpendicular bisectors of ABC, intersect in a point, P. The point where the three perpendicular bisectors of the sides of a triangle intersect is called the circumcenter. Prove that if a point lies on the perpendicular bisector of a line segment, then the point and the endpoints of the line segment are the vertices of an isosceles triangle. Given: P lies on the perpendicular bisector of. RS Prove: RPS is isosceles. P R S 14365C05.pgs 7/10/07 8:41 AM Page 195 Perpendicular Bisector of a Line Segment 195 Proof Statements Reasons 1. P lies on the perpendicular RS bisector of. 2. PR = PS 3. PR > PS 4. RPS is isosceles. 1. Given. 2. If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. (Theorem 5.3b) 3. Segments that have the same mea- sure are congruent. 4. An isosceles triangle is a triangle that has two congru |
ent sides. Exercises Writing About Mathematics 1. Justify the three methods of proving that two lines are perpendicular given in this section. 2. Compare and contrast Example 1 with Corollary 5.1b, “The median from the vertex angle of an isosceles triangle is perpendicular to the base.” Developing Skills 3. If RS ASB is the perpendicular bisector of, prove that ARS BRS. R S B A 5. Polygon ABCD is equilateral (AB = BC = CD = DA). Prove that AC BD other and are perpendicular to each other. and bisect each D C E B A 4. If PR PS and QR QS, prove PQ'RS and RT ST. that R P T Q S 6. Given CED and ADB with ACE BCE and AED BED, prove that is the perpendicular bisector of CED. ADB C E A D B 14365C05.pgs 7/10/07 8:41 AM Page 196 196 Congruence Based on Triangles Applying Skills 7. Prove Theorem 5.2. 8. Prove that if the bisector of an angle of a triangle is perpendicular to the opposite side of the triangle, the triangle is isosceles. 9. A line through one vertex of a triangle intersects the opposite side of the triangle in adjacent angles whose measures are represented by 27 and 15. Is the line perpendicular to the side of the triangle? Justify your answer. 3 2a 1 2a 5-7 BASIC CONSTRUCTIONS A geometric construction is a drawing of a geometric figure done using only a pencil, a compass, and a straightedge, or their equivalents. A straightedge is used to draw a line segment but is not used to measure distance or to determine equal distances. A compass is used to draw circles or arcs of circles to locate points at a fixed distance from given point. The six constructions presented in this section are the basic procedures used for all other constructions. The following postulate allows us to perform these basic constructions: Postulate 5.1 Radii of congruent circles are congruent. Construction 1 Construct a Line Segment Congruent to a Given Line Segment. Given AB Construct CD, a line segment congruent to.AB. With a straightedge, draw a ray, h.CX 2. Open the compass so that the point is on A and the point of the pencil |
is on B. 3. Using the same compass radius, place the point on C and, with the pencil, draw an arc that h CX intersects of intersection D.. Label this point 14365C05.pgs 8/2/07 5:43 PM Page 197 Conclusion CD > AB Proof Since AB and CD are radii of congruent circles, they are congruent. Basic Constructions 197 Construction 2 Construct an Angle Congruent to a Given Angle. Given A Construct EDF BAC A D 1. Draw a ray with endpoint D. C B F E A D 4. Draw h.DF. With A as center, draw an arc that intersects each ray of A. Label the points of intersection B and C. Using the same radius, draw an arc with D as the center that intersects the ray from D at E. 3. With E as the center, draw an arc with radius equal to BC that intersects the arc drawn in step 3. Label the intersection F. Conclusion EDF BAC Proof We used congruent radii to draw, and BC > EF AB > DE DEF ABC by SSS and EDF BAC because they are corresponding parts of congruent triangles.. Therefore, AC > DF, C B F E A D 14365C05.pgs 8/2/07 5:43 PM Page 198 198 Congruence Based on Triangles Construction 3 Given Construct AB g CD Construct the Perpendicular Bisector of a Given Line Segment and the Midpoint of a Given Line Segment.'AB at M, the midpoint of AB.. Open the compass to a radius that is greater than one-half of AB. 2. Place the point of the compass at A and draw an arc above AB arc below and an.AB D 3. Using the same radius, place the point of the compass at B and draw an arc above and an arc AB interbelow AB secting the arcs drawn in step 2. Label the intersections C and D. Conclusion g CD'AB at M, the midpoint of AB. Proof Since they are congruent radii, AD > BD from A and B. If two points are each equidistant. Therefore, C and D are both equidistant AC > BC and from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment (Theorem 5.2). Thus, is the perpen- g CD d |
icular bisector of AB. Finally, M is the point on AB where the perpendicular bisector intersects AB, so AM = BM. By definition, M is the midpoint of AB. M D B 4. Use a straight- edge to draw g CD intersecting AB at M. C A M D B 14365C05.pgs 8/2/07 5:43 PM Page 199 Basic Constructions 199 Construction 4 Bisect a Given Angle. Given ABC Construct h BF, the bisector of ABC. With B as center and any convenient radius, draw an arc that 2. With D as center, draw an arc in the interior of ABC. intersects h BC D and h BA at at E. 3. Using the same 4. Draw h.BF radius, and with E as center, draw an arc that intersects the arc drawn in step 2. Label this intersection F. Conclusion h BF bisects ABC; ABF FBC. Proof We used congruent radii to draw BD > BE and DF > EF. By the reflexive property of congruence, so by SSS, FBD FBE. Therefore, BF > BF ABF FBC because they are corresponding parts of congruent triangles. Then bisects ABC h BF because an angle bisector separates an angle into two congruent angles. B D E A F C Construction 5 will be similar to Construction 4. In Construction 4, any given angle is bisected. In Construction 5, APB is a straight angle that is bisected by the construction. Therefore, APE and BPE are right angles and g PE g.AB ⊥ 14365C05.pgs 8/2/07 5:43 PM Page 200 200 Congruence Based on Triangles Construction 5 Construct a Line Perpendicular to a Given Line Through a Given Point on the Line. Given Point P on g. AB Construct g PE g'AB. With P as center and any convenient radius, draw arcs that interh PB at C and h PA sect at D. 3. Draw g.EP 2. With C and D as centers and a radius greater than that used in step 1, draw arcs intersecting at E. Conclusion g PE g'AB Proof Since points C and D were constructed using congruent radii, CP = PD and P is equidistant to C and D. Similarly, since E was constructed using congruent radii, CE ED, and E is equ |
idistant to C and D. If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendic- ular bisector of the line segment (Theorem 5.2). Therefore, g PE CD. Since CD is the perpendicular bisector of g'AB is a subset of line g AB g PE,. E P C D A B 14365C05.pgs 8/2/07 5:43 PM Page 201 Basic Constructions 201 Construction 6 Construct a Line Perpendicular to a Given Line Through a Point Not on the Given Line. Given Point P not on g. AB Construct g PE g'AB. With P as center and any 2. Open the compass to a convenient radius, draw an arc that intersects two points, C and D. g AB in radius greater than onehalf of CD. With C and D as centers, draw intersecting arcs. Label the point of intersection E. Conclusion g PE g'AB 3. Draw g PE intersecting g AB at F. P F E A C D B Proof Statements Reasons 1. CP > PD, CE > DE 1. Radii of congruent circles are congruent. 2. CP PD, CE DE g PE 3.'CD g PE g'AB 4. 2. Segments that are congruent have the same measure. 3. If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment. (Theorem 5.2) 4. CD is a subset of line g. AB 14365C05.pgs 7/10/07 8:42 AM Page 202 202 Congruence Based on Triangles EXAMPLE 1 Construct the median to AB in ABC. Construction A median of a triangle is a line segment that joins any vertex of the triangle to the midpoint of the opposite side. To construct the median to first find the midpoint of, we must AB AB. 1. Construct the perpendicular bisector of to locate the midpoint. Call the mid- AB point M. 2. Draw. CM Conclusion CM is the median to AB in ABC. A C R M S B Exercises Writing About Mathematics 1. Explain the difference between the altitude of a triangle and the perpendicular bisector of a side of a triangle. 2. Explain how Construction 3 (Construct the perpendicular bisector of a given segment) and Construction 6 (Construct a line perpendicular |
to a given line through a point not on the given line) are alike and how they are different. Developing Skills 3. Given: AB Construct: a. A line segment congruent to. AB b. A line segment whose measure is 2AB. A B c. The perpendicular bisector of d. A line segment whose measure is. AB 11. 2AB 4. Given: A Construct: a. An angle congruent to A. b. An angle whose measure is 2mA. c. The bisector of A. d. An angle whose measure is 21 2m/A. A 14365C05.pgs 7/10/07 8:42 AM Page 203 5. Given: Line segment. ABCD Construct: a. A line segment congruent to. BC Basic Constructions 203 A B C D b. A triangle with sides congruent to, AB BC, and. CD c. An isosceles triangle with the base congruent to AB and with legs congruent to. BC d. An equilateral triangle with sides congruent to. CD 6. Given: A with mA 60. Construct: a. An angle whose measure is 30. b. An angle whose measure is 15. c. An angle whose measure is 45. 7. Given: ABC Construct: a. The median from vertex C. b. The altitude to AB. c. The altitude to BC. A C d. The angle bisector of the triangle from vertex A. A B 8. a. Draw ABC. Construct the three perpendicular bisectors of the sides of ABC. Let P be the point at which the three perpendicular bisectors intersect. b. Is it possible to draw a circle that passes through each of the vertices of the triangle? Explain your answer. Hands-On Activity Compass and straightedge constructions can also be done on the computer by using only the point, line segment, line, and circle creation tools of your geometry software and no other software tools. Working with a partner, use either a compass and straightedge, or geometry software to complete the following constructions: a. A square with side AB. b. An equilateral triangle with side. AB c. 45° angle ABD. d. 30° angle ABD. e. A circle passing through points A, B and C. (Hint: See the proof of Theorem 5.4 or use Theorem 5.3.) A B 14365 |
C05.pgs 8/2/07 5:43 PM Page 204 204 Congruence Based on Triangles CHAPTER SUMMARY Definitions to Know • An altitude of a triangle is a line segment drawn from any vertex of the triangle, perpendicular to and ending in the line that contains the opposite side. • A median of a triangle is a line segment that joins any vertex of the trian- gle to the midpoint of the opposite side. • An angle bisector of a triangle is a line segment that bisects any angle of the triangle and terminates in the side opposite that angle. • The perpendicular bisector of a line segment is a line, a line segment, or a ray that is perpendicular to the line segment at its midpoint. Postulates 5.1 Radii of congruent circles are congruent. Theorems and Corollaries 5.1 If two sides of a triangle are congruent, the angles opposite these sides are congruent. 5.1a The median from the vertex angle of an isosceles triangle bisects the ver- tex angle. 5.1b The median from the vertex angle of an isosceles triangle is perpendicu- lar to the base. 5.1c Every equilateral triangle is equiangular. 5.2 If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment. 5.3 A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the endpoints of the line segment. 5.4 The perpendicular bisectors of the sides of a triangle are concurrent. VOCABULARY 5-1 Altitude of a triangle • Median of a triangle • Angle bisector of a triangle 5-3 Isosceles triangle theorem • Corollary 5-6 Perpendicular bisector of a line segment • Equidistant • Concurrent • Perpendicular bisector concurrence theorem • Circumcenter 5-7 Geometric construction • Straightedge • Compass REVIEW EXERCISES 1. If g LMN'KM, mLMK x y and mKMN 2x y, find the value of x and of y. 2. The bisector of PQR in PQR is QS. If mPQS x 20 and mSQR 5x, find mPQR. 14365C |
05.pgs 8/2/07 5:44 PM Page 205 Review Exercises 205 3. In ABC, CD is both the median and the altitude. If AB 5x 3, AC 2x 8, and BC 3x 5, what is the perimeter of ABC? 4. Angle PQS and angle SQR are a linear pair of angles. If mPQS 5a 15 and mSQR 8a 35, find mPQS and mSQR. 5. Let D be the point at which the three perpendicular bisectors of the sides of equilateral ABC intersect. Prove that ADB, BDC, and CDA are congruent isosceles triangles. side 6. Prove that if the median,, then, to side CD is not congruent to AB is the base of isosceles ABC and g CD AB ABD. Prove that AC 7. AB BC of ABC is not the altitude to. AB is also the base of isosceles is the perpendicular bisector of AB. 8. In ABC, CD is the median to AB mA mB mACB. (Hint: Use Theorem 5.1, “If two sides of a triangle are congruent, the angles opposite these sides are congruent.”). Prove that CD > DB and 9. a. Draw a line, g ADB. Construct g CD'ADB. b. Use ADC to construct ADE such that mADE 45. c. What is the measure of EDC? d. What is the measure of EDB? 10. a. Draw obtuse PQR with the obtuse angle at vertex Q. b. Construct the altitude from vertex P. Exploration As you have noticed, proofs may be completed using a variety of methods. In this activity, you will explore the reasoning of others. 1. Complete a two-column proof of the following: Points L, M, and N separate gruent segments. Point C is not on is an altitude of CLM. Prove that into four conand AB CM CA > CB. AB 2. Cut out each statement and each reason from your proof, omitting the step numbers. 3. Trade proofs with a partner. 4. Attempt to reassemble your partner’s proof. 5. Answer the following questions: C A L M N B a. Were you able to reassemble the |
proof? b. Did the reassembled proof match your partner’s original proof? c. Did you find any components missing or have any components left- over? Why? 14365C05.pgs 7/10/07 8:42 AM Page 206 206 Congruence Based on Triangles CUMULATIVE REVIEW Chapters 1–5 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The symbol g ABC represents (1) a line segment with B between A and C. (2) a line with B the midpoint of (3) a line with B between A and C. (4) a ray with endpoint A. AB. 2. A triangle with no two sides congruent is (1) a right triangle. (2) an equilateral triangle. 3. Opposite rays have (1) no points in common. (2) one point in common. (3) an isosceles triangle. (4) a scalene triangle. (3) two points in common. (4) all points in common. 4. The equality a 1 1 a is an illustration of (1) the commutative property of addition. (2) the additive inverse property. (3) the multiplicative identity property. (4) the closure property of addition. 5. The solution set of the equation 1.5x 7 0.25x 8 is (1) 120 (2) 12 (3) 11 (4) 1.2 6. What is the inverse of the statement “When spiders weave their webs by noon, fine weather is coming soon”? (1) When spiders do not weave their webs by noon, fine weather is not coming soon. (2) When fine weather is coming soon, then spiders weave their webs by noon. (3) When fine weather is not coming soon, spiders do not weave their webs by noon. (4) When spiders weave their webs by noon, fine weather is not coming soon. 7. If ABC and CBD are a linear pair of angles, then they must be (1) congruent angles. (2) complementary angles. (3) supplementary angles. (4) vertical angles. 14365C05.pgs 7/10/07 8:42 AM Page 207 Chapter Summary 207 8. Which of the following is not an abbreviation for a postulate that is used to prove triangles congruent? |
(1) SSS (2) SAS (3) ASA (4) SSA 9. If the statement “If two angles are right angles, then they are congruent” is true, which of the following statements must also be true? (1) If two angles are not right angles, then they are not congruent. (2) If two angles are congruent, then they are right angles. (3) If two angles are not congruent, then they are not right angles. (4) Two angles are congruent only if they are right angles. 10. If is a line, which of the following may be false? g ABC (1) B is on (2). AC AB 1 BC 5 AC (3) B is between A and C. (4) B is the midpoint of AC. Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Angle PQS and angle SQR are a linear pair of angles. If mPQS 3a 18 and mSQR = 7a 2, find the measure of each angle of the linear pair. 12. Give a reason for each step in the solution of the given equation. 5(4 x) 32 x 20 5x 32 x 20 5x x 32 x x 20 6x 32 0 20 6x 32 6x 12 x 2 Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 14365C05.pgs 7/10/07 8:42 AM Page 208 208 Congruence Based on Triangles 13. Given:, ABF is the AE > BF supplement of A, and AB > CD. Prove: AEC BFD E F 14. Given: AB > CB point on of ABC. h BD and E is any, the bisector Prove: AE > CE C E D A B C D B A Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, |
diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Given: Point P is not on ABCD and P PB PC. Prove: ABP DCP A B C D 16. Prove that if AC and BD are perpendicular bisectors of each other, quadrilateral ABCD is equilateral (AB BC CD DA). 14365C06.pgs 7/12/07 2:57 PM Page 209 TRANSFORMATIONS AND THE COORDINATE PLANE In our study of mathematics, we are accustomed to representing an equation as a line or a curve. This blending of geometry and algebra was not always familiar to mathematicians. In the seventeenth century, René Descartes (1596–1650), a French philosopher and mathematician, applied algebraic principles and methods to geometry. This blending of algebra and geometry is known as coordinate geometry or analytic geometry. Pierre de Fermat (1601–1665) independently developed analytic geometry before Descartes but Descartes was the first to publish his work. Descartes showed that a curve in a plane could be completely determined if the distances of its points from two fixed perpendicular lines were known. We call these distances the Cartesian coordinates of the points; thus giving Descartes’ name to the system that he developed. CHAPTER 6 CHAPTER TABLE OF CONTENTS 6-1 The Coordinates of a Point in a Plane 6-2 Line Reflections 6-3 Line Reflections in the Coordinate Plane 6-4 Point Reflections in the Coordinate Plane 6-5 Translations in the Coordinate Plane 6-6 Rotations in the Coordinate Plane 6-7 Glide Reflections 6-8 Dilations in the Coordinate Plane 6-9 Transformations as Functions Chapter Summary Vocabulary Review Exercises Cumulative Review 209 14365C06.pgs 7/12/07 2:57 PM Page 210 210 Transformations and the Coordinate Plane 6-1 THE COORDINATES OF A POINT IN A PLANE y 3 2 1 –3 –2 –1 –1 –2 –3 O 1 2 3 x In this chapter, we will review what you know about the coordinate plane and you will study transformations and symmetry, which are common elements in nature, art, and architecture. Two intersecting lines determine a plane. The coordinate plane is determined by a horizontal line, the x-axis, and |
a vertical line, the y-axis, which are perpendicular and intersect at a point called the origin. Every point on a plane can be described by two numbers, called the coordinates of the point, usually written as an ordered pair. The first number in the pair, called the x-coordinate or the abscissa, is the distance from the point to the y-axis. The second number, the y-coordinate or the ordinate is the distance from the point to the x-axis. In general, the coordinates of a point are represented as (x, y). Point O, the origin, has the coordinates (0, 0). We will accept the following postulates of the coordinate plane. Postulate 6.1 Two points are on the same horizontal line if and only if they have the same y-coordinates. Postulate 6.2 The length of a horizontal line segment is the absolute value of the difference of the x-coordinates. Postulate 6.3 Two points are on the same vertical line if and only if they have the same x-coordinate. Postulate 6.4 The length of a vertical line segment is the absolute value of the difference of the y-coordinates. Postulate 6.5 Each vertical line is perpendicular to each horizontal line. Locating a Point in the Coordinate Plane An ordered pair of signed numbers uniquely determines the location of a point in the plane. 14365C06.pgs 7/12/07 2:57 PM Page 211 The Coordinates of a Point in a Plane 211 Procedure To locate a point in the coordinate plane: 1. From the origin, move along the x-axis the number of units given by the xcoordinate. Move to the right if the number is positive or to the left if the number is negative. If the x-coordinate is 0, there is no movement along the x-axis. 2. Then, from the point on the x-axis, move parallel to the y-axis the number of units given by the y-coordinate. Move up if the number is positive or down if the number is negative. If the y-coordinate is 0, there is no movement in the y direction. For example, to locate the point A(3, 4), from O, move 3 units to the left along the x-axis, then 4 units down, parallel to the y-axis. y O x A Finding the Coordinates of a Point in a Plane The location of |
a point in the coordinate plane uniquely determines the coordinates of the point. Procedure To find the coordinates of a point: 1. From the point, move along a vertical line to the x-axis.The number on the x-axis is the x-coordinate of the point. 2. From the point, move along a horizontal line to the y-axis.The number on the y-axis is the y-coordinate of the point. For example, from point R, move in the vertical direction to 5 on the x-axis and in the horizontal direction to 6 on the y-axis. The coordinates of R are (5, 6). Note: The coordinates of a point are often referred to as rectangular coordinates(5, –6) 14365C06.pgs 7/12/07 2:57 PM Page 212 212 Transformations and the Coordinate Plane Graphing Polygons A quadrilateral (a four-sided polygon) can be represented in the coordinate plane by locating its vertices and then drawing the sides connecting the vertices in order. y 1 A(3, 2) O 1 x The graph shows the quadrilateral ABCD. The vertices are A(3, 2), B(3, 2), C(3, 2) and D(3, 2). 1. Points A and B have the same y-coordinate and are on the same horizontal line. 2. Points C and D have the same y-coordinate and are on the same horizontal line. 3. Points B and C have the same x-coordinate and are on the same vertical line. 4. Points A and D have the same x-coordinate and are on the same vertical line. 5. Every vertical line is perpendicular to every horizontal line. 6. Perpendicular lines are lines that intersect to form right angles. Each angle of the quadrilateral is a right angle: mA mB mC mD 90 From the graph, we can find the dimensions of this quadrilateral. To find AB and CD, we can count the number of units from A to B or from C to D. AB CD 6 Points on the same horizontal line have the same y-coordinate. Therefore, we can also find AB and CD by subtracting their x-coordinates. AB CD 3 (3) 3 3 6 To find BC and DA, we can count the number of units from B to C or from D to A. BC DA 4 |
Points on the same vertical line have the same x-coordinate. Therefore, we can find BC and DA by subtracting their y-coordinates. BC DA 2 (2) 2 2 4 14365C06.pgs 7/12/07 2:57 PM Page 213 The Coordinates of a Point in a Plane 213 EXAMPLE 1 Graph the following points: A(4, 1), B(1, 5), C(2,1). Then draw ABC and find its area. Solution The graph shows ABC. To find the area of the triangle, we need to know the lengths of the base and of the altitude drawn to that base. The base of ABC is AC, a horizontal line segment. AC 4 (2) 4 2 6 y B(1, 5) O D(1, 1) A(4, 1) 1 x The vertical line segment drawn from B perpendicular to AC is the altitude. BD BD 5 1 4 Area 1 2(AC)(BD) 1 2(6)(4) 12 Answer The area of ABC is 12 square units. Exercises Writing About Mathematics 1. Mark is drawing isosceles right triangle ABC on the coordinate plane. He locates points A(2, 4) and C(5, 4). He wants the right angle to be C. What must be the coordinates of point B? Explain how you found your answer. / 2. Phyllis graphed the points D(3, 0), E(0, 5), F(2, 0), and G(0, 4) on the coordinate plane and joined the points in order. Explain how Phyllis can find the area of this polygon. Developing Skills In 3–12: a. Graph the points and connect them with straight lines in order, forming a polygon. b. Find the area of the polygon. 3. A(1, 1), B(8, 1), C(1, 5) 5. C(8, 1), A(9, 3), L(4, 3), F(3, 1) 4. P(0, 0), Q(5, 0), R(5, 4), S(0, 4) 6. H(4, 0), O(0, 0), M(0, 4), E(4, 4) 14365C06.pgs 7/12/07 2:57 PM Page 214 214 Transformations and the Coordinate Plane 7. H |
(5, 3), E(5, 3), N(2, 0) 9. B(3, 2), A(2, 2), R(2, 2), N(3, 2) 11. R(4, 2), A(0, 2), M(0, 7) 8. F(5, 1), A(5, 5), R(0, 5), M(2, 1) 10. P(3, 0), O(0, 0), N(2, 2), D(1, 2) 12. M(1, 1), I(3, 1), L(3, 3), K(1, 3) 13. Graph points A(1, 1), B(5, 1), and C(5, 4). What must be the coordinates of point D if ABCD is a quadrilateral with four right angles? 14. Graph points P(1, 4) and Q(2, 4). What are the coordinates of R and S if PQRS is a quadrilateral with four right angles and four congruent sides? (Two answers are possible.) 15. a. Graph points S(3, 0), T(0, 4), A(3, 0), and R(0, 4), and draw the polygon STAR. b. Find the area of STAR by adding the areas of the triangles into which the axes sepa- rate the polygon. 16. a. Graph points P(2, 0), L(1, 1), A(1, 1), N(2, 0). E(1, 1), and T(1, 1), and draw the hexagon PLANET. b. Find the area of PLANET. (Hint: Use two vertical lines to separate the hexagon into parts.) 6-2 LINE REFLECTIONS It is often possible to see the objects along the shore of a body of water reflected in the water. If a picture of such a scene is folded, the objects can be made to coincide with their images. Each point of the reflection is an image point of the corresponding point of the object. The line along which the picture is folded is the line of reflection, and the correspondence between the object points and the image points is called a line reflection. This common experience is used in mathematics to study congruent figures. If the figure at the left were folded along line k, ABC would coincide with ABC. Line k is the line |
of reflection, point A corresponds to point A (in symbols, A → A) and point B corresponds to point B (B → B). Point C is a fixed point because it is a point on the line of reflection. In other words, C corresponds to itself (C → C). Under a reflection in line k, then, ABC corresponds to ABC (ABC → ABC). Each of the points A, B, and C is called a preimage and each of the points A, B, and C is called an image. C B B A A k 14365C06.pgs 7/12/07 2:57 PM Page 215 DEFINITION A transformation is a one-to-one correspondence between two sets of points, S and S, such that every point in set S corresponds to one and only one point in set S, called its image, and every point in S is the image of one and only one point in S, called its preimage. S S Line Reflections 215 The sets S and S can be the same set and that set of points is frequently the set of points in a plane. For example, let S be the set of points in the coordinate plane. Let the image of (x, y), a point in S, be (2 x, y), a point in S. Under this transformation: (0, 1) → (2 0, 1) (2, 1) (4, 3) → (2 (4), 3) (2, 3) (5, 1) → (2 5, 1) (3, 1) Every image (2 x, y) in S is a point of the coordinate plane, that is, a point of S. Therefore, S S. P P P k DEFINITION A reflection in line k is a transformation in a plane such that: 1. If point P is not on k, then the image of P is P where k is the perpendicular bisector of PPr. 2. If point P is on k, the image of P is P. From the examples that we have seen, it appears that the size and shape of the image is the same as the size and shape of the preimage. We want to prove that this is true. Theorem 6.1 Under a line reflection, distance is preserved. Given Under a reflection in line k, the image of A is A and the image of B is B. Prove AB AB B D A C Proof We will prove |
this theorem by using the definition of a reflection and SAS to show that BDC BDC and that using the fact that if two angles are congruent, their complements are congruent (Theorem 4.4), we can show that ACB ACB. From this last statement, we can conclude that AB AB. BC > BrC. Then, k 14365C06.pgs 7/12/07 2:57 PM Page 216 216 Transformations and the Coordinate Plane (1) Let the image of A be A and the image of B be B under reflection in k. Let C be the and D be the midpoint of midpoint of BBr. Points C and D are on k, since k is the perpendicular bisector of. and of AAr AAr BBr (2) since D is the midpoint of BD > BrD, BDC BDC since perpendicular lines intersect to form right angles, and CD > CD BDC BDC by SAS. by the reflexive property. Thus, BBr (3) From step 2, we can conclude that and BCD BCD since BC > BrC corresponding parts of congruent triangles are congruent. (4) Since k is the perpendicular bisector of, ACD and ACD are right angles. AAr Thus, ACB and BCD are complementary angles. Similarly, ACB and BCD are complementary angles. If two angles are congruent, their complements are congruent. Since BCD and BCD were shown to be congruent in step 3, their complements are congruent: ACB ACB. (5) By SAS, ACB ACB. Since corresponding parts of congruent trian. Therefore, AB = AB by the definition of AB > ArBr gles are congruent, congruent segments. A similar proof can be given when A and B are on opposite sides of the line k. The proof is left to the student. (See exercise 11.) Since distance is preserved under a line reflection, we can prove that the image of a triangle is a congruent triangle and that angle measure, collinearity, and midpoint are also preserved. For each of the following corollaries, the image of A is A, the image of B is B, and the image of C is C. Therefore, in the diagram, ABC ABC by SSS. We will use this fact to prove the following coroll |
aries: B C D A M k M 14365C06.pgs 7/12/07 2:57 PM Page 217 Corollary 6.1a Under a line reflection, angle measure is preserved. Line Reflections 217 Proof: We found that ABC ABC. Therefore, ABC ABC because the angles are corresponding parts of congruent triangles. Corollary 6.1b Under a line reflection, collinearity is preserved. D A B C k Proof: Let D be a point on AB whose image is D. Since distance is preserved: AD AD DB DB A, D, and B are collinear, D is between A and B, and AD DB AB. By substitution, AD DB AB. AB = AB If D were not on, AD DB AB because a straight line is the shortest distance between two points. But by substitution, AD DB AB. Therefore, A, D and B are collinear and D is between A and B. ArBr Corollary 6.1c Under a line reflection, midpoint is preserved. A M M B C k Proof: Let M be the midpoint of preserved under a line reflection, AM AM, and MC MC. AC Since M is the midpoint of AC, AM MC and, by the substitution postu- and M the image of M. Since distance is late, AM MC. Therefore, M is the midpoint of ArCr, that is, midpoint is preserved under a line reflection. We can summarize Theorem 6.1 and its corollaries in the following statement: Under a line reflection, distance, angle measure, collinearity, and midpoint are preserved. We use rk as a symbol for the image under a reflection in line k. For example, rk(A) B means rk(ABC) ABC means “The image of A under a reflection in line k is B.” “The image of ABC under a reflection in line k is ABC.” 14365C06.pgs 7/12/07 2:57 PM Page 218 218 Transformations and the Coordinate Plane EXAMPLE 1 If rk(CD) CrDr, construct CrDr. Construction 1. Construct the perpendicular line from C to k. Let the point of intersection be M. 2. Construct the perpendicular line from D to k. Let the point of intersection be N. 3. Construct point C on such that CM MC and point D on g such that |
DN ND. DN g CM 4. Draw CrDr. Line Symmetry We know that the altitude to the base of an isosceles triangle is also the median and the perpendicular bisector of the base. If we imagine that isosceles triangle ABC, shown at the left, is folded along the perpendicular bisector of the base so that A falls on C, the line along which it folds, k, is a reflection line. Every point of the triangle has as its image a point of the triangle. Points B and D are fixed points because they are points of the line of reflection. Thus, under the line reflection in k: 1. All points of ABC are reflected so that C → A 2. The sides of ABC are reflected; that is, A → C E → F that the legs of an isosceles triangle are congruent. Also, ing that the base is its own image AB S CB, a statement verifying AC S CA, show- B E F A D C k 14365C06.pgs 7/12/07 2:57 PM Page 219 Line Reflections 219 3. The angles of ABC are reflected; that is, BAD → BCD, a statement verifying that the base angles of an isosceles triangle are congruent. Also, ABC → CBA, showing that the vertex angle is its own image. We can note some properties of a line reflection by considering the reflec- tion of isosceles triangle ABC in line k: 1. Distance is preserved (unchanged). AB S CB and AB CB AD S CD and AD CD 2. Angle measure is preserved. BAD → BCD and mBAD mBCD BDA → BDC and mBDA mBDC 3. The line of reflection is the perpendicular bisector of every segment join- ing a point to its image. The line of reflection, bisector of AC. g BD, is the perpendicular 4. A figure is always congruent to its image: ABC CBA. In nature, in art, and in industry, many forms have a pleasing, attractive appearance because of a balanced arrangement of their parts. We say that such forms have symmetry. In each of the figures above, there is a line on which the figure could be folded so that the parts of the figure on opposite sides of the line would coincide. If we think of that line as a line of reflection, each point of the figure has as its image a point |
of the figure. This line of reflection is a line of symmetry, or an axis of symmetry, and the figure has line symmetry. DEFINITION A figure has line symmetry when the figure is its own image under a line reflection. 14365C06.pgs 7/12/07 2:57 PM Page 220 220 Transformations and the Coordinate Plane It is possible for a figure to have more than one axis of symmetry. In the square g XY is an axis of symme- PQRS at the right, g VW try and is a second axis of symmetry. The diagonals, PR and QS, are also segments of axes of symmetry. Lines of symmetry may be found for some letters and for some words, as shown at the left. BD Not every figure, however, has line symmetry. If the parallelogram ABCD at the right is reflected in the line that contains the, the image of A is A and the diagonal image of C is C. Points A and C, however, are not points of the original parallelogram. The image of parallelogram ABCD under a is ABCD. Therefore, g. BD ABCD is not symmetric with respect to reflection in g BD We have used the line that contains the diagonal as a line of reflection, but note that it is not a line of symmetry. There is no line along which the parallelogram can be folded so that the image of every point of the parallelogram will be a point of the parallelogram under a reflection in that line. BD EXAMPLE 2 How many lines of symmetry does the letter H have? Solution The horizontal line through the crossbar is a line of symmetry. The vertical line midway between the vertical segments is also a line of symmetry. Answer The letter H has two lines of symmetry. Exercises Writing About Mathematics 1. Explain how a number line illustrates a one-to-one correspondence between the set of points on a line and the set of real numbers. 2. Is the correspondence (x, y) → (2, y) a transformation? Explain why or why not. 14365C06.pgs 7/12/07 2:57 PM Page 221 Line Reflections 221 Developing Skills 3. If rk(PQR) PQR and PQR is isosceles with PQ QR, prove that PQR is isosceles. 4. If rk(LMN) LMN and LMN is |
a right triangle with mN 90, prove that LMN is a right triangle. In 5–9, under a reflection in line k, the image of A is A, the image of B is B, the image of C is C, and the image of D is D. 5. Is ABC ABC? Justify your answer. 6. If AC'BC, is ArCr'BrCr? Justify your answer? 7. The midpoint of AB is M. If the image of M is M, is M the midpoint of ArBr? Justify your answer. 8. If A, M, and B lie on a line, do A, M, and B lie on a line? Justify your answer. 9. Which point lies on the line of reflection? Justify your answer. 10. A triangle, RST, has line symmetry with respect to the altitude from S but does not have line symmetry with respect to the altitudes from R and T. What kind of a triangle is RST? Justify your answer. Applying Skills 11. Write the proof of Theorem 6.1 for the case when points A and B are on opposite sides of line k. A k B 12. A baseball diamond is in the shape of a rhombus with sides 90 feet in length. Describe the lines of symmetry of a baseball diamond in terms of home plate and the three bases. 13. Print three letters that have line symmetry with respect to only one line and draw the line of symmetry. 14. Print three letters that have line symmetry with respect to exactly two lines and draw the lines of symmetry. 14365C06.pgs 7/12/07 2:57 PM Page 222 222 Transformations and the Coordinate Plane 6-3 LINE REFLECTIONS IN THE COORDINATE PLANE We can apply the definition of a line reflection to points in the coordinate plane. Reflection in the y-axis In the figure, ABC is reflected in the y-axis. Its image under the reflection is ABC. From the figure, we see that: A(1, 2) → A(1, 2) B(3, 4) → B(3, 4) C(1, 5) → C(1, 5) C(1, 5) B(3, 4) A(1, 2) y 1 C(1, 5) B(3, 4) A(1, 2) For each point and its image |
under a reflection in the y-axis, the y-coordinate of the image is the same as the y-coordinate of the point; the x-coordinate of the image is the opposite of the x-coordinate of the point. Note that for a reflection in the y-axis, the image of (1, 2) is (1, 2) and the image of (1, 2) is (1, 2). O x 1 A reflection in the y-axis can be designated as ry-axis. For example, if the image of (1, 2) is (1, 2) under a reflection in the y-axis, we can write: ry-axis(1, 2) (1, 2) From these examples, we form a general rule that can be proven as a theorem. Theorem 6.2 Under a reflection in the y-axis, the image of P(a, b) is P(a, b). Given A reflection in the y-axis. Prove The image of P(a, b) under a reflection in the y-axis is P(a, b). P(a, b) y O Q(0, b) P(a, b) x Proof By the definition of a reflection in a line, a point P is the image of P under a reflection in a given line if and only if the line is the perpendicular bisector of. Therefore, we can prove that P is the image of P under a reflection in PPr the y-axis by showing that the y-axis is the perpendicular bisector of PPr. (1) The y-axis is perpendicular to. The line of reflection, the y-axis, is a PPr vertical line. P(a, b) and P(a, b) have the same y-coordinates. is a segment of a horizontal line because two points are on the same horizontal line if and only if they have the same y-coordinates. Every vertical line is perpendicular to every horizontal line. Therefore, the y-axis is perpendicular to.PPr PPr 14365C06.pgs 7/12/07 2:57 PM Page 223 Line Reflections in the Coordinate Plane 223 PPr (2) The y-axis bisects. Let Q be the point at which y-axis. The x-coordinate of every point on the y-axis is 0. The length of a horizontal line segment |
is the absolute value of the difference of the x-coordinates of the endpoints. intersects the PPr PQ a 0 a and PQ a 0 = a Since PQ PQ, Q is the midpoint of Steps 1 and 2 prove that if P has the coordinates (a, b) and P has the coor, and therefore, dinates (a, b), the y-axis is the perpendicular bisector of the image of P(a, b) is P(a, b). or the y-axis bisects PPr PPr PPr. Reflection in the x-axis In the figure, ABC is reflected in the x-axis. Its image under the reflection is ABC. From the figure, we see that: A(1, 2) → A(1, 2) B(3, 4) → B(3, 4) C(1, 5) → C(1, 5) For each point and its image under a reflection in the x-axis, the x-coordinate of the image is the same as the x-coordinate of the point; the y-coordinate of the image is the opposite of the y-coordinate of the point. Note that for a reflection in the x-axis, the image of (1, 2) is (1, 2) and the image of (1, 2) is (1, 2). y C(1, 5) B(3, 4) A(1, 2) 1 x 1 O A reflection in the x-axis can be designated as rx-axis. For example, if the image of (1, 2) is (1, 2) under a reflection in the x-axis, we can write: rx-axis(1, 2) (1, 2) From these examples, we form a general rule that can be proven as a theorem. Theorem 6.3 Under a reflection in the x-axis, the image of P(a, b) is P(a, b). The proof follows the same general pattern as that for a reflection in the y-axis. Prove that the x-axis is the perpendicular bisector of. The proof is left to the student. (See exercise 18.) PPr y O P(a, b) x Q(a, 0) P(a, b) 14365C06.pgs 7/12/07 2:57 PM Page 224 224 |
Transformations and the Coordinate Plane EXAMPLE 1 On graph paper: a. Locate A(3, 1). b. Locate A, the image of A under a reflection in the y-axis, and write its coordinates. c. Locate A, the image of A under a reflection in the x-axis, and write its coordinates. Answers y 1 1O x Reflection in the Line y = x In the figure, ABC is reflected in the line y x. Its image under the reflection is ABC. From the figure, we see that: y C(1, 5) B(3, 4) A(1, 2) → A(2, 1) B(3, 4) → B(4, 3) C(1, 5) → C(5, 1) For each point and its image under a reflection in the line y = x, the x-coordinate of the image is the y-coordinate of the point; the y-coordinate of the image is the x-coordinate of the point. Note that for a reflection in the line y x, the image of (1, 2) is (2, 1) and the image of (2, 1) is (1, 2). B(4, 3) C(5, 1) x A(1, 2) A(2, 1) O 11 A reflection in the line y x can be designated as ry = x. For example, if the image of (1, 2) is (2, 1) under a reflection in y x, we can write: From these examples, we form a general rule that can be proven as a theorem. ry = x(1, 2) (2, 1) Theorem 6.4 Under a reflection in the line y = x, the image of P(a, b) is P(b, a). Given A reflection in the line whose equation is y x. Prove The image of P(a, b) is P(b, a). 14365C06.pgs 7/12/07 2:57 PM Page 225 Line Reflections in the Coordinate Plane 225 Proof For every point on the line y x, the xcoordinate and the y-coordinate are equal. Locate the points Q(a, a) and R(b, b) on the line y x. If two points have the same x-coordinate, then |
the distance between them is the absolute value of their y-coordinates and if two points have the same y-coordinate, then the distance between them is the absolute value of their x-coordinates. PQ b a and PQ b a Therefore, Q is equidistant from P and P. y P(a, b) R(b, b) Q(a, a) P(b, a) x PR b a and PR = b a Therefore, R is equidistant from P and P. If two points are each equidistant from the endpoints of a line segment, they lie on the perpendicular bisector of the line segment (Theorem 5.3). Therefore,. By RQ the definition of a line reflection across the line y x, the image of P(a, b) is P(b, a)., which is a subset of the line y x, is the perpendicular bisector of PPr EXAMPLE 2 The vertices of DEF are D(2, 2), E(0, 3), and F(2, 0). a. Draw DEF on graph paper. b. Draw the image of DEF under a reflection in the line whose equation is y = x. Answers y E(3, 0) D(2, 2) 1 O 1 F(2, 0) x Exercises Writing About Mathematics 1. When finding the distance from P(a, b) to Q(a, a), Allison wrote PQ b a and Jacob wrote PQ a b. Explain why both are correct. 2. The image of point A is in the third quadrant under a reflection in the y-axis and in the first quadrant under a reflection in the x-axis. In what quadrant is the image of A under a reflection in the line y x? 14365C06.pgs 7/12/07 2:57 PM Page 226 226 Transformations and the Coordinate Plane Developing Skills In 3–7: a. On graph paper, locate each point and its image under rx-axis. b. Write the coordinates of the image point. 3. (2, 5) 4. (1, 3) 5. (2, 3) 6. (2, 4) 7. (0, 2) In 8–12: a. On graph paper, locate each point and its image under ry-axis. b. Write the coordinates of the image point. 8. |
(3, 5) 9. (1, 4) 10. (2, 3) 11. (2, 3) 12. (1, 0) In 13–17: a. On graph paper, locate each point and its image under ryx. b. Write the coordinates of the image point. 13. (3, 5) 14. (3, 5) 15. (4, 2) 16. (1, 5) 17. (2, 2) Applying Skills 18. Prove Theorem 6.3, “Under a reflection in the x-axis, the image of P(a, b) is P(a, b).” 19. When the points A(4, 0), B(0, 4), C(4, 0) and D(0, 4) are connected in order, square ABCD is drawn. a. Show that the line y = x is a line of symmetry for the square. b. Show that the y-axis is a line of symmetry for the square. 20. Show that the y-axis is not a line of symmetry for the rectangle whose vertices are E(0, 3), F(5, 3), G(5, 3), and H(0, 3). 21. Write the equation of two lines that are lines of symmetry for the rectangle whose vertices are E(0, 3), F(6, 3), G(6, 3), and H(0, 3). Hands-On Activity 1 In this activity, you will learn how to construct a reflection in a line using a compass and a straightedge, or geometry software. (Note: Compass and straightedge constructions can also be done on the computer by using only the point, line segment, line, and circle creation tools of your geometry software and no other software tools.) STEP 1. Draw a line segment. Label the endpoints A and B. Draw a reflection line k. STEP 2. Construct line l perpendicular to line k through point A. Let M be the point where lines l and k intersect. STEP 3. Construct line segment along line l. Using point M as the center, AM draw a circle with radius equal to AM. Let A be the point where the circle intersects line l on the ray that is the opposite ray of congruent to h. MA ArM STEP 4. Repeat steps 2 and 3 for point B in order to construct B. STEP 5. Draw ArBr. Result |
: ArBr is the image of AB under a reflection in line k. 14365C06.pgs 7/12/07 2:57 PM Page 227 For each figure, construct the reflection in the given line. Point Reflections in the Coordinate Plane 227 AB a. Segment with vertices A(4, 2) and B(2, 4), and line k through points C(2, 1) and D(0, 5). b. Angle EFG with vertices E(4, 3), F(1, 6), and G(6, 6), and line l through points H(1, 3) and I(3, 1). c. Triangle JKL with vertices J(2, 2), K(5, 1), and L(1, 4), and line p through points M(7, 3) and N(2, 0). PQ d. Segments and RS points T(0, 3) and U(3, 0). with vertices P(3, 4), Q(3, 2), R(1, 0), and S(1, 6), and line q through Hands-On Activity 2 In each figure, one is the image of the other under a reflection in a line. For each figure, construct the reflection line using a compass and a straightedge. a. A A b. B A c. Triangle ABC with vertices A(1, 4), B(4, 3), and (2, 8), and triangle ABC with vertices A(3, 2), B(2, 3), C(7, 3). d. Triangle ABC with vertices A(0, 4), B(7, 1), and C(6, 6), and triangle ABC with vertices A(7, 3), B(4, 4), C(9, 3). 6-4 POINT REFLECTIONS IN THE COORDINATE PLANE The figure at the right illustrates another type of reflection, a reflection in a point. In the figure, ABC is the image of ABC under a reflection in point P. If a line segment is drawn connecting any point to its image, then the point of reflection is the midpoint of that segment. In the figure: • Point A is on midpoint of, AP PA, and P is the g AP AAr. g BP g CP • Point B is on, BP PB, and P is the midpoint of • Point C is on |
, CP PC, and P is the midpoint of.CCr A C P B BBr. DEFINITION A point reflection in P is a transformation of the plane such that: 1. If point A is not point P, then the image of A is A and P the midpoint of 2. The point P is its own image. AAr. 14365C06.pgs 7/12/07 2:57 PM Page 228 228 Transformations and the Coordinate Plane Properties of Point Reflections Looking at triangles ABC and ABC and point of reflection P, we observe some properties of a point reflection: A C P B 1. Distance is preserved: AB AB. 2. Angle measure is preserved: mACB mACB. 3. A figure is always congruent to its image. 4. Collinearity is preserved. The image of any point on AB is a point on ArBr. 5. Midpoint is preserved. The image of the midpoint of AB is the midpoint of ArBr. We can state the first property as a theorem and the remaining properties as corollaries to this theorem. Theorem 6.5 Under a point reflection, distance is preserved. Given Under a reflection in point P, the image of A is A and the image of B is B. Prove AB AB Proof Since APB and APB are vertical angles, APB APB. In a point reflection, if a point X is not point P, then the image of X is X XXr and P the midpoint of. Therefore, P is the AAr BBr BP > BPr and midpoint of both. By SAS, APB APB and AP > PAr and AB AB., and A B P B A The case when either point A or B is point P is left to the student. (See exercise 9.) Corollary 6.5a Under a point reflection, angle measure is preserved. Corollary 6.5b Under a point reflection, collinearity is preserved. Corollary 6.5c Under a point reflection, midpoint is preserved. 14365C06.pgs 7/12/07 2:57 PM Page 229 Point Reflections in the Coordinate Plane 229 We can prove that angle measure, collinearity, and midpoint are preserved using the same proofs that we used to prove the corollaries of Theorem 6.1. (See exercises 10–12.) Theorem 6. |
5 and its corollaries can be summarized in the following statement. Under a point reflection, distance, angle measure, collinearity, and midpoint are preserved. We use RP as a symbol for the image under a reflection in point P. For example, RP(A) B means “The image of A under a reflection in point P is B.” R(1, 2)(A) A means “The image of A under a reflection in point (1, 2) is A.” Point Symmetry DEFINITION A A figure has point symmetry if the figure is its own image under a reflection in a point. P B A circle is the most common example of a figure with point symmetry. Let P be the center of a circle, A be any point on the circle, and B be the other point intersect the circle. Since every point on a circle is equidistant from g AP at which the center, PA PB, P is the midpoint of image of A under a reflection in P. AB and B, a point on the circle, is the Other examples of figures that have point symmetry are letters such as S and N and numbers such as 8. Point Reflection in the Coordinate Plane In the coordinate plane, the origin is the most common point that is used to used to define a point reflection. In the diagram, points A(3, 5) and B(2, 4) are reflected in the origin. The coordinates of A, the image of A, are (3, 5) and the coordinates of B, the image of B, are (2, 4). These examples suggest the following theorem. A(3, 5) B(–2, 4) y 1 O 1 x B(2, 4) 14365C06.pgs 7/12/07 2:57 PM Page 230 230 Transformations and the Coordinate Plane Theorem 6.6 Under a reflection in the origin, the image of P(a, b) is P(a, b). Given A reflection in the origin. Prove Under a reflection in the origin, the image of P(a, b) is P(a, b). y P(a, b) x O B(a, 0) Proof Let P be the image of P(a, b) under a reflection in the origin, O. Then: OP OP and OP > OPr Let B(a, 0) be the point of intersection of |
the x-axis and a vertical line through P. Then: OB 0 a a Let B be the point (a, 0). Then: OB 0 (a) a OB OB and OB > OBr POB POB because vertical angles are congruent. Therefore, POB POB by SAS. In particular, P PB PB 0 b b Since OBP is a right angle, OBP is a right angle and is a vertical line. Therefore, P has the same x-coordinate as B and is b units in the opposite direction from the x-axis as P. The coordinates of P are (a, b). PrBr Note: a and b are the opposites of a and b. The image of (4, 2) is (4, 2) and the image of (5, 1) is (5, 1). We can symbolize the reflection in the origin as RO. Therefore, we can write: RO(a, b) (a, b) EXAMPLE 1 a. What are the coordinates of B, the image of A(3, 2) under a reflection in the origin? b. What are the coordinates of C, the image of A(3, 2) under a reflection in the x-axis? 14365C06.pgs 7/12/07 2:57 PM Page 231 Point Reflections in the Coordinate Plane 231 c. What are the coordinates of D, the image of C under a reflection in the y-axis? d. Does a reflection in the origin give the same result as a reflection in the x-axis followed by a reflection in the y-axis? Justify your answer. Solution a. Since RO(a, b) (a, b), RO(3, 2) (3, 2). The coordinates of B are (3, 2). Answer b. Since rx-axis (a, b) (a, b), rx-axis (3, 2) (3, 2). The coordinates of C are (3, 2). Answer c. Since ry-axis (a, b) (a, b), ry-axis (3, 2) (3, 2). The coordinates of D are (3, 2). Answer d. For the point A(3, 2), a reflection in the origin gives the same result as a reflection in the x-axis followed by a reflection in the y-axis. In general, RO(a, b |
) (a, b) while rx-axis(a, b) (a, b) ry-axis(a, b) (a, b) Therefore, a reflection in the origin gives the same result as a reflection in the x-axis followed by a reflection in the y-axis. Answer Exercises Writing About Mathematics 1. Ada said if the image of AB under a reflection in point P is ArBr, then the image of ArBr under a reflection in point P is AB. Do you agree with Ada? Justify your answer. 2. Lines l and m intersect at P. ArBr is the image of AB under a reflection in line l and is the image of P? Justify your answer. ArBr under a reflection in line m. Is the result the same if AB ArrBrr is reflected in Developing Skills In 3–8, give the coordinates of the image of each point under RO. 3. (1, 5) 4. (2, 4) 5. (1, 0) 6. (0, 3) 7. (6, 6) 8. (1, 5) 14365C06.pgs 7/12/07 2:57 PM Page 232 232 Transformations and the Coordinate Plane Applying Skills 9. Write a proof of Theorem 6.5 for the case when either point A or B is point P, that is, when the reflection is in one of the points. 10. Prove Corollary 6.5a, “Under a point reflection, angle measure is preserved.” 11. Prove Corollary 6.5b, “Under a point reflection, collinearity is preserved.” 12. Prove Corollary 6.5c, “Under a point reflection, midpoint is preserved.” 13. The letters S and N have point symmetry. Print 5 other letters that have point symmetry. 14. Show that the quadrilateral with vertices P(5, 0), Q(0, 5), R(5, 0), and S(0, 5) has point symmetry with respect to the origin. 15. a. What is the image of A(2, 6) under a reflection in P(2, 0)? b. What is the image of B(3, 6) under a reflection in P(2, 0)? c. Is the reflection in the point P(2, 0) the |
same as the reflection in the x-axis? Justify your answer. 16. a. What is the image of A(4, 4) under a reflection in the point P(4, 2)? b. What is the image of C(1, 2) under a reflection in the point P(4, 2)? c. The point D(2, 0) lies on the segment AB. Does the image of D lie on the image of AB? Justify your answer. 6-5 TRANSLATIONS IN THE COORDINATE PLANE It is often useful or necessary to move objects from one place to another. If we move a table from one place in the room to another, the bottom of each leg moves the same distance in the same direction. DEFINITION A translation is a transformation of the plane that moves every point in the plane the same distance in the same direction. C If ABC is the image of ABC under a translation, AA BB CC. It appears that the size and shape of the figure are unchanged, so that ABC ABC. Thus, under a translation, as with a reflection, a figure is congruent to its image. The following example in the coordinate plane shows that this is true. In the coordinate plane, the distance is given in terms of horizontal distance (change in the x-coordinates) and vertical distance (change in the y-coordinates). A B 14365C06.pgs 7/12/07 2:57 PM Page 233 Translations in the Coordinate Plane 233 In the figure, DEF is translated by moving every point 4 units to the right and 5 units down. From the figure, we see that: y F(1, 6) E(4, 2) D(1, 2) F(5, 1) 1 x 1 O D(5, 3) E(8, 3) D(1, 2) → D(5, 3) E(4, 2) → E(8, 3) F(1, 6) → F(5, 1) • DE DrEr and DE 1 4 3 are horizontal segments. DE 5 8 3 • DF and DrFr are vertical segments. DF 2 6 DF 3 1 4 4 • FDE and FDE are right angles. Therefore, DEF DEF by SAS. This translation moves every point 4 units to the right (4) and 5 units down (5). 1. The x-coordinate of the image is 4 more than the x- |
coordinate of the point: x → x 4 2. The y-coordinate of the image is 5 less than the y-coordinate of the point: From this example, we form a general rule: y → y 5 DEFINITION A translation of a units in the horizontal direction and b units in the vertical direction is a transformation of the plane such that the image of P(x, y) is P(x a, y b). Note: If the translation moves a point to the right, a is positive; if it moves a point to the left, a is negative; if the translation moves a point up, b is positive; if it moves a point down, b is negative. 14365C06.pgs 7/12/07 2:57 PM Page 234 234 Transformations and the Coordinate Plane Theorem 6.7 Under a translation, distance is preserved. Given A translation in which the image b) of A(x1, y1) is A(x1 a, y1 and the image of B(x2, y2) is B(x2 a, y2 b). Prove AB = AB y B(x2 a, y2 b) B(x2, y2) A(x1 a, y1 b) C(x2 a, y1 b) C(x2, y1) A(x1, y1) Proof Locate C(x2, y1), a point on the same horizontal line as A and the same vertical line as B. Then by definition, the image of C is O x Thus: C(x2 a, y1 b). BC 5 y2 2 y1 BrCr 5 (y2 1 b) 2 (y1 1 b) 5 y2 2 y1 1 b 2 b 5 y2 2 y1 AC 5 x1 2 x2 ArCr 5 (x1 1 a) 2 (x2 1 a) 5 x1 2 x2 1 a 2 a 5 x1 2 x2 Since horizontal and vertical lines are perpendicular, BCA and BCA are both right angles. By SAS, ABC ABC. Then, AB and AB are the equal measures of the corresponding sides of congruent triangles. When we have proved that distance is preserved, we can prove that angle measure, collinearity, and midpoint are preserved. The proofs are similar to the proofs of the corollaries of Theorem 6.1 and |
are left to the student. (See exercise 10.) Corollary 6.7a Under a translation, angle measure is preserved. Corollary 6.7b Under a translation, collinearity is preserved. Corollary 6.7c Under a translation, midpoint is preserved. 14365C06.pgs 7/31/07 1:21 PM Page 235 Translations in the Coordinate Plane 235 We can write Theorem 6.7 and its corollaries as a single statement. Under a translation, distance, angle measure, collinearity, and midpoint are preserved. Let Ta,b be the symbol for a translation of a units in the horizontal direction and b units in the vertical direction. We can write: Ta,b(x, y) (x a, y b) EXAMPLE 1 The coordinates of the vertices of ABC are A(3, 3), B(3, 2), and C(6, 1). a. Find the coordinates of the vertices of ABC, the image of ABC under T5,3. b. Sketch ABC and ABC on graph paper. Solution a. Under the given translation, every point moves 5 units to the left and 3 units up. b. y B(2, 5) A(3, 3) → A(3 5, 3 3) A(2, 0) B(3, 2) → B(3 5, 2 3) B(2, 5) C(6, 1) → C(6 5, 1 3) C(1, 4) C(1, 4) B(3, 2) C(6, 1) 1 A(2, 0) O 1 x A(3, 3) Translational Symmetry DEFINITION A figure has translational symmetry if the image of every point of the figure is a point of the figure. Patterns used for decorative purposes such as wallpaper or borders on clothing often appear to have translational symmetry. True translational symmetry would be possible, however, only if the pattern could repeat without end. 14365C06.pgs 7/12/07 2:57 PM Page 236 236 Transformations and the Coordinate Plane Exercises Writing About Mathematics 1. Explain why there can be no fixed points under a translation other than T0,0. 2. Hunter said that if A is the image of A under a reflection in the y-axis and A is the image of A under a |
reflection in the line x 3, then A is the image of A under the translation (x, y) → (x 6, y). a. Do you agree with Hunter when A is a point with an x-coordinate greater than or equal to 3? Justify your answer. b. Do you agree with Hunter when A is a point with an x-coordinate greater than 0 but less than 3? Justify your answer. c. Do you agree with Hunter when A is a point with a negative x-coordinate? Justify your answer. Developing Skills 3. The diagram consists of nine congruent rectangles. Under a A B C D translation, the image of A is G. Find the image of each of the given points under the same translation. a. J b. B c. I d. F e. a. On graph paper, draw and label ABC, whose vertices have the coordinates A(1, 2), B(6, 3), and C(4, 6). b. Under the translation P(x, y) → P(x 5, y 3), the image of ABC is ABC. Find the coordinates of A, B, and C. c. On the same graph drawn in part a, draw and label ABC. 5. a. On graph paper, draw and label ABC if the coordinates of A are (2, 2), the coordi- nates of B are (2, 0), and the coordinates of C are (3, 3). b. On the same graph, draw and label ABC, the image of ABC under the transla- tion T4,7. c. Give the coordinates of the vertices of ABC. 6. If the rule of a translation is written (x, y) → (x a, y b), what are the values of a and b for a translation where every point moves 6 units to the right on a graph? 7. In a translation, every point moves 4 units down. Write a rule for this translation in the form (x, y) → (x a, y b). 14365C06.pgs 7/12/07 2:57 PM Page 237 Translations in the Coordinate Plane 237 8. The coordinates of ABC are A(2, 1), B(4, 1), and C(5, 5). a. On graph paper, draw and label ABC. b. Write a rule for the translation in which the image of A is |
C(5, 5). c. Use the rule from part b to find the coordinates of B, the image of B, and C, the image of C, under this translation. d. On the graph drawn in part a, draw and label CBC, the image of ABC. 9. The coordinates of the vertices of ABC are A(0, 1), B(2, 1), and C(3, 3). a. On graph paper, draw and label ABC. b. Under a translation, the image of C is B(2, 1). Find the coordinates of A, the image of A, and of B, the image of B, under this same translation. c. On the graph drawn in part a, draw and label ABB, the image of ABC. d. How many points, if any, are fixed points under this translation? Applying Skills 10. Prove the corollaries of Theorem 6.7. a. Corollary 6.7a, “Under a translation, angle measure is preserved.” b. Corollary 6.7b, “Under a translation, collinearity is preserved.” c. Corollary 6.7c, “Under a translation, midpoint is preserved.” (Hint: See the proofs of the corollaries of Theorem 6.1 on page 217.) 11. The coordinates of the vertices of LMN are L(6, 0), M(2, 0), and N(2, 2). a. Draw LMN on graph paper. b. Find the coordinates of the vertices of LMN, the image of LMN under a reflection in the line x 1, and draw LMN on the graph drawn in a. c. Find the coordinates of the vertices of LMN, the image of LMN under a reflec- tion in the line x = 4, and draw LMN on the graph drawn in a. d. Find the coordinates of the vertices of PQR, the image of LMN under the translation T10,0. e. What is the relationship between LMN and PQR? 12. The coordinates of the vertices of DEF are D(2, 3), E(1, 3), and F(0, 0). a. Draw DEF on graph paper. b. Find the coordinates of the vertices of DEF, the image of DEF under a reflection in the line |
y 0 (the x-axis), and draw DEF on the graph drawn in a. c. Find the coordinates of the vertices of DEF, the image of DEF under a reflec- tion in the line y 3, and draw DEF on the graph drawn in a. d. Find the coordinates of the vertices of RST, the image of DEF under the translation T0,6. e. What is the relationship between DEF and RST? 14365C06.pgs 7/12/07 2:57 PM Page 238 238 Transformations and the Coordinate Plane 13. The coordinates of the vertices of ABC are A(1, 2), B(5, 2), and C(4, 5). a. Draw ABC on graph paper. b. Find the coordinates of the vertices of ABC, the image of ABC under a reflection in the line y 0 (the x-axis), and draw ABC on the graph drawn in a. c. Find the coordinates of the vertices of ABC, the image of ABC under a reflec- tion in the line y 3, and draw ABC on the graph drawn in a. d. Is there a translation, Ta,b, such that ABC is the image of ABC? If so, what are the values of a and b? 14. In exercises, 11, 12, and 13, is there a relationship between the distance between the two lines of reflection and the x-values and y-values in the translation? 15. The coordinates of the vertices of ABC are A(1, 2), B(5, 2), and C(4, 5). a. Draw ABC on graph paper. b. Find the coordinates of the vertices of ABC, the image of ABC under a reflection in any horizontal line, and draw ABC on the graph drawn in a. c. Find the coordinates of the vertices of ABC, the image of ABC under a reflection in the line that is 3 units below the line of reflection that you used in b, and draw ABC on the graph drawn in a. d. What is the translation Ta,b such that ABC is the image of ABC? Is this the same translation that you found in exercise 13? 6-6 ROTATIONS IN THE COORDINATE PLANE Think of what happens to all of the points of a wheel as the wheel is turned. Except for the fixed point in the center, every point moves through a part of a circle, or arc, so |
that the position of each point is changed by a rotation of the same number of degrees. DEFINITION A rotation is a transformation of a plane about a fixed point P through an angle of d degrees such that: 1. For A, a point that is not the fixed point P, if the image of A is A, then PA = PA and mAPA d. 2. The image of the center of rotation P is P. In the figure, P is the center of rotation. If A is rotated about P to A, and B is rotated the same number of degrees to B, then mAPA mBPB. Since P is the center of rotation, PA PA and PB PB, and mAPA mBPA mAPB mBPB mBPA mAPB B A P 14365C06.pgs 7/12/07 2:57 PM Page 239 Rotations in the Coordinate Plane 239 Therefore, mAPB = mAPB and APB APB by SAS. Because corresponding parts of congruent triangles are congruent and equal in measure, AB AB, that is, that distance is preserved under a rotation. We have just proved the following theorem: Theorem 6.8 Distance is preserved under a rotation about a fixed point. As we have seen when studying other transformations, when distance is pre- served, angle measure, collinearity, and midpoint are also preserved. Under a rotation about a fixed point, distance, angle measure, collinearity, and midpoint are preserved. We use RP,d as a symbol for the image under a rotation of d degrees about point P. For example, the statement “RO,30°(A) B” can be read as “the image of A under a rotation of 30° degrees about the origin is B.” A rotation in the counterclockwise direction is called a positive rotation. For instance, B is the image of A under a rotation of 30° about P. A rotation in the clockwise direction is called a negative rotation. For instance, C is the image of A under a rotation of 45° about P. B A P C Rotational Symmetry DEFINITION A figure is said to have rotational symmetry if the figure is its own image under a rotation and the center of rotation is the only fixed point. Many letters, as well as designs in the shapes of wheels, stars, and polygons, have rotational symmetry. When |
a figure has rotational symmetry under a rotation of do, we can rotate the figure by do to an image that is an identical figure. Each figure shown below has rotational symmetry. 14365C06.pgs 7/12/07 2:57 PM Page 240 240 Transformations and the Coordinate Plane Any regular polygon (a polygon with all sides congruent and all angles congruent) has rotational sym3608, metry. When regular pentagon ABCDE is rotated 5 or 72°, about its center, the image of every point of the figure is a point of the figure. Under this rotation The figure would also have rotational symmetry if it were rotated through a multiple of 72° (144°, 216°, or 288°). If it were rotated through 360°, every point would be its own image. Since this is true for every figure, we do not usually consider a 360° rotation as rotational symmetry. Rotations in the Coordinate Plane The most common rotation in the coordinate plane is a quarter turn about the origin, that is, a counterclockwise rotation of 90° about the origin. In the diagram, the vertices of right triangle ABC are A(0,0), B(3, 4) and C(3, 0). When rotated 90° about the origin, A remains fixed because it is the center of rotation. The image of C, which is on the x-axis and 3 units from the origin, is C(0, 3) on the y-axis and 3 units from the origin. Since is a vertical line 4 units long, its image is a horizontal line 4 units long and to the left of the y-axis. Therefore, the image of B is B(4, 3). Notice that the x-coordinate of B is the negative of the y-coordinate of B and the y-coordinate of B is the x-coordinate of B. CB y B(3, 4) A(0, 0) C(3, 0) x The point B(3, 4) and its image B(4, 3) in the above example suggest a rule for the coordinates of the image of any point P(x, y) under a counterclockwise rotation of 90° about the origin. Theorem 6.9 Under a counterclockwise rotation of 90° about the origin, the image of P(a, b) is P(–b, a). Proof: We |
will prove this theorem by using a rectangle with opposite vertices at the origin and at P. Note that in quadrants I and II, when b is positive, b is negative, and in quadrants III and IV, when b is negative, b is positive. 14365C06.pgs 7/12/07 2:57 PM Page 241 S(0, b) y P(a, b) x O R(a, 0) Rotations in the Coordinate Plane 241 y P(a, b) S(0, b) x R(a, 0) O y y R(a, 0) O x P(a, b) S(0, b) R(a, 0) x P(a, b) O S(0, b) Let P(a, b) be any point not on an axis and O(0, 0) be the origin. Let R(a, 0) be the point on the x-axis on the same vertical line as P and S(0, b) be the point on the y-axis on the same horizontal line as P. Therefore, PR b 0 b and PS a 0 a. P(b, a) y R(0, a) S(0, b) P(a, b) c b x S(b, 0) O a R(a, 0) c OR Under a rotation of 90° about the origin, the image of ORPS is ORPS. Then, OR OR and mROR 90. This means that since is a horizontal ORr is a vertical segment with segment, the same length. Therefore, since R is on the x-axis, R is on the y-axis and the coordinates of R are (0, a). Similarly, since S is on the y-axis, S is on the x-axis and the coordinates of S are (b, 0). Point P is on the same horizontal line as R and therefore has the same y-coordinate as R, and P is on the same vertical line as S and has the same x-coordinate as S. Therefore, the coordinates of P are (b, a). The statement of Theorem 6.9 may be written as: RO,90°(x, y) (y, x) or R90°(x, y) (y, x) When the point that is the center of rotation is not named, it is understood to be O |
, the origin. Note: The symbol R is used to designate both a point reflection and a rotation. 1. When the symbol R is followed by a letter that designates a point, it repre- sents a reflection in that point. 2. When the symbol R is followed by both a letter that designates a point and the number of degrees, it represents a rotation of the given number of degrees about the given point. 3. When the symbol R is followed by the number of degrees, it represents a rotation of the given number of degrees about the origin. 14365C06.pgs 7/12/07 2:57 PM Page 242 242 Transformations and the Coordinate Plane EXAMPLE 1 Point P is at the center of equilateral triangle ABC (the point at which the perpendicular bisectors of the sides intersect) so that: PA PB PC and mAPB mBPC mCPA A P Under a rotation about P for which the image of A is B, find: B C a. The number of degrees in the rotation. b. The image of B. c. The image of d. The image of CAB.. CA Answers a. 360o 3 5 120o b. C c. AB d. ABC EXAMPLE 2 What are the coordinates of the image of P(2, 3) under R90°? Solution The image of (x, y) is (y, x). Therefore, the image of (2, 3) is (3, 2). Answer (3, 2) Exercises Writing About Mathematics 1. A point in the coordinate plane is rotated 180° about the origin by rotating the point counterclockwise 90° and then rotating the image counterclockwise 90°. a. Choose several points in the coordinate plane and find their images under a rotation of 180°. What is the image of P(x, y) under this rotation? b. For what other transformation does P(x, y) have the same image? c. Is a 180° rotation about the origin equivalent to the transformation found in part b? 2. Let Q be the image of P(x, y) under a clockwise rotation of 90° about the origin and R be the image of Q under a clockwise rotation of 90° about the origin. For what two different transformations is R the image of P? 14365C06.pgs 7/12/07 2:57 PM Page 243 Developing Skills For 3 and |
4, refer to the figure at the right. 3. What is the image of each of the given points under R90°? a. A e. H b. B f. J c. C g. K d. G h. L 4. What is the image of each of the given points under R–90°? a. A b. B c. C d. G Glide Reflections 243 y C B A O x LK J D EF G H I e. H h. L 5. The vertices of rectangle ABCD are A(2, 1), B(5, 1), C(5, 4), and D(2, 4). g. K f. J a. What are the coordinates of the vertices of ABCD, the image of ABCD under a counterclockwise rotation of 90° about the origin? b. Is ABCD a rectangle? c. Find the coordinates of the midpoint of AB and of ArBr. Is the midpoint of ArBr the image of the midpoint of AB under this rotation? 6. a. What are the coordinates of Q, the image of P(1, 3) under a counterclockwise rotation of 90° about the origin? b. What are the coordinates of R, the image of Q under a counterclockwise rotation of 90° about the origin? c. What are the coordinates of S, the image of R under a counterclockwise rotation of 90° about the origin? d. What are the coordinates of P, the image of P under a clockwise rotation of 90° about the origin? e. Explain why S and P are the same point. 6-7 GLIDE REFLECTIONS When two transformations are performed, one following the other, we have a composition of transformations. The first transformation produces an image and the second transformation is performed on that image. One such composition that occurs frequently is the composition of a line reflection and a translation. DEFINITION A glide reflection is a composition of transformations of the plane that consists of a line reflection and a translation in the direction of the line of reflection performed in either order. 14365C06.pgs 7/12/07 2:57 PM Page 244 244 Transformations and the Coordinate Plane y C The vertices of ABC are A(1, 2), B(5, 3), and C(3, 4). Under a reflection in the y-axis, the image of ABC is ABC whose |
vertices are A(1, 2), B(5, 3), and C(3, 4). Under the translation T0,–4, the image of ABC is ABC whose vertices are A(1, 2), B(5, 1), and C(3, 0). Under a glide reflection, the image of ABC is ABC. Note that the line of reflection, the y-axis, is a vertical line. The translation is in the vertical direction because the x-coordinate of the translation is 0. Distance is preserved under a line reflection and under a translation: ABC ABC ABC. Distance is preserved under a glide reflection. We have just proved the following theorem: O A B 1 1 x Theorem 6.10 Under a glide reflection, distance is preserved. As we have seen with other transformations, when distance is preserved, angle measure, collinearity, and midpoint are also preserved. Therefore, we may make the following statement: Under a glide reflection, distance, angle measure, collinearity, and midpoint are preserved. In this chapter, we have studied five transformations: line reflection, point reflection, translation, rotation, and glide reflection. Each of these transformations is called an isometry because it preserves distance. DEFINITION An isometry is a transformation that preserves distance. EXAMPLE 1 The vertices of PQR are P(2, 1), Q(4, 1), and R(4, 3). a. Find PQR, the image of PQR under ryx followed by T3,3. b. Find PQR, the image of PQR under T3,3 followed by ryx. c. Are PQR and PQR the same triangle? d. Are ryx followed by T3,3 and T3,3 followed by ryx the same glide reflection? Explain. e. Write a rule for this glide reflection. 14365C06.pgs 7/12/07 2:57 PM Page 245 Solution T3,3(y, x) (y 3, x 3) a. ryx(x, y) (y, x) T3,3(1, 2) (2, 1) ryx(2, 1) (1, 2) T3,3(1, 4) (2, 1) ryx(4, 1) (1, 4) ryx(4, 3) (3, 4) T |
3,3(3, 4) (0, 1) The vertices of PQR are P(2, 1), Q(2, 1), and R(0, 1). Answer Glide Reflections 245 y R Q x P b. T3,3(x, y) (x 3, y 3) T3,3(2, 1) (1, 2) T3,3(4, 1) (1, 2) T3,3(4, 3) (1, 0) The vertices of PQR are P (2, 1), Q(2, 1), and R(0, 1). Answer ryx(x 3, y 3) (y – 3, x 3) ryx(1, 2) (2, 1) ryx(1, 2) (2, 1) ryx(1, 0) (0, 1. PQR and PQR are the same triangle. Answer d. The image of (x, y) under ryx followed by T3,3 is (x, y) → (y, x) → (y 3, x 3) The image of (x, y) under T3,3 followed by ryx is (x, y) → (x 3, y 3) → (y 3, x 3) ryx followed by T3,3 and T3,3 followed by ryx are the same glide reflection. Answer e. (x, y) → (y 3, x 3) Answer EXAMPLE 2 Is a reflection in the y-axis followed by the translation T5,5 a glide reflection? Solution The y-axis is a vertical line. The translation T5,5 is not a translation in the verti- cal direction. Therefore, a reflection in the y-axis followed by the translation T5,5 is not a glide reflection. 14365C06.pgs 7/12/07 2:57 PM Page 246 246 Transformations and the Coordinate Plane Exercises Writing About Mathematics 1. Does a glide reflection have any fixed points? Justify your answer. 2. In a glide reflection, the line reflection and the translation can be done in either order. Is the composition of any line reflection and any translation always the same in either order? Justify your answer. Developing Skills In 3–8, a. find the coordinates of the image of the triangle whose vertices are |
A(1, 1), B(5, 4), and C(3, 5) under the given composition of transformations. b. Sketch ABC and its image. c. Explain why the given composition of transformations is or is not a glide reflection. d. Write the coordinates of the image of (a, b) under the given composition of transformations. 3. A reflection in the x-axis followed by a translation of 4 units to the right. 4. A reflection in the y-axis followed by a translation of 6 units down. 5. T5,0 followed by rx-axis 7. T3,3 followed by ry=x Applying Skills 6. T3,3 followed by ry-axis 8. R90 followed by T1,2 9. The point A(4, 5) is the image of A under a glide reflection that consists of a reflection in the x-axis followed by the translation T2,0. What are the coordinates of A? 10. Under a glide reflection, the image of A(2, 4) is A(2, 7) and the image of B(3, 7) is B(3, 4). a. If the line of reflection is a vertical line, write an equation for the line reflection and a rule for the translation. b. What are the coordinates of C, the image of C(0, 7), under the same glide reflection? 11. Under a glide reflection, the image of A(3, 4) is A(1, 4) and the image of B(5, 5) is B(1, 5). a. If the line of reflection is a horizontal line, write a rule for the line reflection and a rule for the translation. b. What are the coordinates of C, the image of C(4, 2), under the same glide reflection? 12. a. For what transformation or transformations are there no fixed points? b. For what transformation or transformations is there exactly one fixed point? c. For what transformation or transformations are there infinitely many fixed points? 14365C06.pgs 7/12/07 2:57 PM Page 247 Dilations in the Coordinate Plane 247 6-8 DILATIONS IN THE COORDINATE PLANE In this chapter, we have learned about transformations in the plane that are isometries, that is, transformations that preserve distance. There is another transformation in the plane that preserves angle measure but not distance. This transformation is |
a dilation. For example, in the coordinate plane, a dilation of 2 with center at the origin will stretch each ray by a factor of 2. If the image of A is A, h then A is a point on OA and OA 2OA. y O A A x DEFINITION A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr and OP kOP. 3. When k is negative and the image of P is P, then rays and OP kOP. are the same ray h OP and h OPr are opposite Note: In step 3, when k is negative, k is positive. In the coordinate plane, the center of dilation is usually the origin. If the center of dilation is not the origin, the coordinates of the center will be given. In the coordinate plane, under a dilation of k with the center at the origin: P(x, y) → P(kx, ky) or Dk(x, y) (kx, ky) For example, the image of ABC is ABC under a dilation of. The vertices of ABC are A(2, 6), B(6, 4), and C(4, 0). Under a dilation of, the rule is 1 2 1 2 A (x, y) 5 2x, 1 1 D1 2y B 2 A(2, 6) → A(1, 3) B(6, 4) → B(3, 2) C(4, 0) → C(2, 0 14365C06.pgs 7/12/07 2:57 PM Page 248 248 Transformations and the Coordinate Plane By the definition of a dilation, distance is not preserved. We will prove in Chapter 12 that angle measure, collinearity, and midpoint are preserved. Under a dilation about a fixed point, distance is not preserved. EXAMPLE 1 The coordinates of parallelogram EFGH are E(0, 0), F(3, 0), G(4, 2), and H(1, 2). Under D3, the image of EFGH is EFGH. a. Find the coordinates of the vertices of EFGH. b. Let M be the midpoint of EF. Find the |
coordinates of M and of M, the image of M. Verify that the midpoint is preserved. Solution a. D3(x, y) (3x, 3y). Therefore, E(0, 0), F(9, 0), G(12, 6), and H(3, 6). Answer b. Since E and F lie on the x-axis, EF is the absolute value of the difference of their x-coordinates. EF 3 0 3. 11 Therefore, the midpoint, M, is 2 from E or from F on the x-axis. The 5 coordinates of M are units, 0 0 1 11 11 2, 0 A Since E and F lie on the x-axis, EF is the absolute value of the difference of their x-coordinates. EF 9 0 9. Therefore, the midpoint, 41 units from E or from F on the x-axis. The coordinates of M are M, is 2 0 1 41 2, 0 B 11 D3 2, 0 A preserved. Answer. Therefore, the image of M is M and midpoint is 41. 2, 0 B 41 2, 0 A 5 5 A B B A EXAMPLE 2 Find the coordinates of P, the image of P(4, 5) under the composition of transformations: D2 followed by rx-axis. Solution The dilation is to be performed first: D2(4, 5) (8, 10). Then perform the reflection, using the result of the dilation: rx-axis(8, 10) (8, 10). Answer P (8, 10) 14365C06.pgs 7/12/07 2:57 PM Page 249 Dilations in the Coordinate Plane 249 Exercises Writing About Mathematics 1. Let ABC be the image of ABC under a dilation of k. When k 1, how do the lengths of the sides of ABC compare with the lengths of the corresponding sides of ABC? Is a dilation an isometry? 2. Let ABC be the image of ABC under a dilation of k. When 0 k 1, how do the lengths of the sides of ABC compare with the lengths of the corresponding sides of ABC? Developing Skills In 3–6, use the rule (x, y) → (4x, 4y) to find the coordinates of the image of each given point. 3. (3, 7) 5. (2, 0) 4. (4, 2) |
6. (1, 9) In 7–10, find the coordinates of the image of each given point under D5. 7. (2, 2) 9. (3, 5) 8. (1, 10) 10. (0, 4) In 11–14, each given point is the image under D2. Find the coordinates of each preimage. 11. (6, 2) 13. (6, 5) 12. (4, 0) 14. (10, 7) In 15–20, find the coordinates of the image of each given point under the given composition of transformations. 15. D4(1, 5) followed by ry-axis 17. T2,1(4, 2) followed by D1 2 19. T1,1(4, 2) followed by D2 16. D2(3, 2) followed by R90° 18. D3(5, 1) followed by rx-axis 20. D2(6, 3) followed by ryx In 21–24, each transformation is the composition of a dilation and a reflection in either the x-axis or the y-axis. In each case, write a rule for composition of transformations for which the image of A is A. 21. A(2, 5) → A(4, 10) 23. A(10, 4) → A(5, 2) Applying Skills 22. A(3, 1) → A(21, 7) 24. A(20, 8) → A(5, 2) 25. The vertices of rectangle ABCD are A(1, 1), B(3, 1), C(3, 3), and D(1, 3). a. Find the coordinates of the vertices of ABCD, the image of ABCD under D4. b. Show that ABCD is a rectangle. 14365C06.pgs 7/12/07 2:57 PM Page 250 250 Transformations and the Coordinate Plane 26. Show that when k 0, a dilation of k with center at the origin followed by a reflection in the origin is the same as a dilation of k with center at the origin. 27. a. Draw ABC, whose vertices are A(2, 3), B(4, 3), and C(4, 6). b. Using the same set of axes, graph ABC, the image of ABC under a dilation of |
3. c. Using ABC and its image ABC, show that distance is not preserved under the dilation. Hands-On Activity In this activity, we will verify that angle measure, collinearity, and midpoint are preserved under a dilation. Using geometry software, or a pencil, ruler and a protractor, draw the pentagon A(1, 2), B(4, 2), C(6, 4), D(2, 8), E(1, 6), and point P(4, 6) on. For each given dilation: a. Measure the corresponding angles. Do they appear to be congruent? b. Does the image of P appear? c. Does the image of the midpoint of to be on the image of appear to be the midCD point of the image of? AE (1) D3 (2) D–3 D1 2 D21 AE CD (4) (3) 2 6-9 TRANSFORMATIONS AS FUNCTIONS A function is a set of ordered pairs in which no two pairs have the same first element. For example, the equation y x 5 describes the set of ordered pairs of real numbers in which the second element, y, is 5 more than the first element, x. The set of first elements is the domain of the function and the set of second ele3, 17 2 ments is the range. Some pairs of this function are (3, 2), (2.7, 2.3),, 3 B (0, 5), and (1, 6). Since the set of real numbers is infinite, the set of pairs of this function cannot be listed completely and are more correctly defined when they are described. Each of the following notations describes the function in which the second element is 5 more than the first. A f {(x, y) y x 5} f: x → x 5 y x 5 f(x) x 5 Note that f(x) and y each name the second element of the ordered pair. Each of the transformations defined in this chapter is a one-to-one function. It is a set of ordered pairs in which the first element of each pair is a point of the plane and the second element is the image of that point under a transformation. For each point in the plane, there is one and only one image. For example, compare the function f(x) x 5 and a line reflection. 14365C06.pgs 7/12/ |
07 2:57 PM Page 251 f: A one-to-one algebraic function f(x) x 5 y f(x) x 5 1 O 1 x Transformations as Functions 251 rm: A reflection in line m A B D C E m 1. For every x in the domain there is one and only one y in the range. For example: f(3) 8 f(0) 5 f(2) 3 2. Every f(x) or y in the range corresponds to one and only one value of x in the range. For example: If f(x) 4, then x 1. If f(x) 9, then x 4. 3. For the function f: domain {real numbers} range {real numbers} 1. For every point in the plane, there is one and only one image of that point in the plane. For example: rm(A) A rm(C) C rm(D) D 2. Every point is the image of one and only one preimage. For example: The point B has the preimage B. The point E has itself as the preimage. The point C has the preimage C. 3. For the function rm: domain {points in the plane} range {points in the plane} Composition of Transformations We defined the composition of transformations as a combination of two transformations in which the first transformation produces an image and the second transformation is performed on that image. For example, to indicate that A is the image of A under a reflection in the line y = x followed by the translation T2,0, we can write T2,0(ryx(A)) A. 14365C06.pgs 7/12/07 2:57 PM Page 252 252 Transformations and the Coordinate Plane If the coordinates of A are (2, 5), we start with A and perform the transformations from right to left as indicated by the parentheses, evaluating what is in parentheses first. ryx(2, 5) (5, 2) followed by T2,0(5, 2) (7, 2) or T2,0(ry 5 x(2, 5)) 5 T2,0(5, 2) 5 (7, 2) A small raised circle is another way of indicating a composition of transfor- mations. T2,0(ryx (A)) A can also be written as + ry 5 x(A) 5 Ar. Again |
, we start with the coordinates of A and move from right to left. Find the image of A under the reflection and then, using the coordinates of that image, find the coordinates under the translation. T2,0 Orientation The figures below show the images of ABC under a point reflection, a translation, and a rotation. In each figure, the vertices, when traced from A to B to C are in the clockwise direction, called the orientation of the points. The vertices of the images, when traced in the same order, from A to B to C are also in the clockwise direction. Therefore, we say that under a point reflection, a translation, or a rotation, orientation is unchanged Point Reflection Translation P Rotation DEFINITION A direct isometry is a transformation that preserves distance and orientation. Point reflection, rotation, and translation are direct isometries. 14365C06.pgs 7/12/07 2:57 PM Page 253 The figure at the right shows ABC and its image under a line reflection. In this figure, the vertices, when traced from A to B to C, are again in the clockwise direction. The vertices of the images, when traced in the same order, from A to B to C, are in the counterclockwise direction. Therefore, under a line reflection, orientation is changed or reversed. Transformations as Functions 253 B A C Line Reflection DEFINITION An opposite isometry is a transformation that preserves distance but changes the order or orientation from clockwise to counterclockwise or from counterclockwise to clockwise. A line reflection is an opposite isometry. EXAMPLE 1 The vertices of RST are R(1,1), S(6, 3), and T(2, 5). a. Sketch RST. b. Find the coordinates of the vertices of RST under the composition rx-axis + ry-axis and sketch RST. c. Is the composition a direct isometry? d. For what single transformation is the image the same as that of rx-axis Solution a. + ry-axis? y 1 R O1 T b. rx-axis + ry-axis( S x rx-axis + ry-axis( rx-axis + ry-axis( 1, 1) rx-axis(1, 1) (1, 1) 6, 3) rx-axis(6, 3) (6, |
3) 2, 5) rx-axis(2, 5) (2, 5) c. A line reflection is an opposite d. isometry. The composition of two opposite isometries is a direct isometry. This composition is a direct isometry. + ry-axis( x, y) (x, y) rx-axis and RO(x, y) (–x, y). The composition of a reflection in the y-axis followed by a reflection in the x-axis is a reflection in the origin. 14365C06.pgs 7/12/07 2:57 PM Page 254 254 Transformations and the Coordinate Plane EXAMPLE 2 The transformation (x, y) → (y 3, x 2) is the composition of what two transformations? Solution Under a rotation of 90° about the origin, (x, y) → (y, x). When this rotation is followed by the translation T3,–2(y, x) (y 3, x 2). Answer T3,22 + R908(x,y) T3,2(y, x) (y 3, x 2) This solution is not unique. Alternative Solution Under the translation T2,3(x, y) (x 2, y 3). When this translation is followed by the rotation R90°(x 2, y 3) ((y 3), x 2) (y 3, x 2). Answer R908 + T22,23(x, y) R90°(x 2, y 3) (y 3, x 2) Exercises Writing About Mathematics 1. Owen said that since a reflection in the x-axis followed by a reflection in the y-axis has the same result as a reflection in the y-axis followed by a reflection in the x-axis, the composition of line reflections is a commutative operation. Do you agree with Owen? Justify your answer. 2. Tyler said that the composition of an even number of opposite isometries is a direct isometry and that the composition of an odd number of opposite isometries is an opposite isometry. Do you agree with Tyler? Justify your answer. Developing Skills In 3–11, find the image of A(3, 2) under the given composition of transformations. 3. 6. rx-axis R908 ry-axis + ry-axis + RO + ry 5 x 4. |
7. ry-axis T21,4 RO + rx-axis + ry 5 x + T3,22 9. 12. A reflection in the line y x followed by a rotation of 90° about the origin is equivalent to 10. 11. what single transformation? 13. A rotation of 90° about the origin followed by another rotation of 90° about the origin is equivalent to what single transformation? 5. 8. RO R908 T22,4 + rx-axis + R908 + D22 14365C06.pgs 8/2/07 5:45 PM Page 255 14. The vertices of DEF are D(3, 2), E(5, 5), and F(4, 1). Chapter Summary 255 a. If T23,0 + rx-axis(nDEF) T23,0 DEF, find the coordinates of the vertices of DEF. + rx-axis is an example of what type of transformation? b. The composition c. Is the composition T23,0 + rx-axis a direct or an opposite isometry? Applying Skills 15. Prove that under a reflection in the line y x, A(a, b) → A(b, a). Suggested plan: Let B(a, a) and C(b, b) be points on the line y x. Show that the line y x is the perpendicular bisector of AAr. 16. Prove that the composition of a reflection in the line y x followed by a reflection in the line y x is a reflection in the origin. 17. Prove that the composition of a reflection in the line y x followed by a reflection in the line y x is a reflection in the origin. CHAPTER SUMMARY Definitions to Know • A transformation is a one-to-one correspondence between two sets of points, S and S, when every point in S corresponds to one and only one point in S called its image, and every point in S is the image of one and only one point in S called its preimage. • A reflection in line k is a transformation in a plane such that: 1. If point A is not on k, then the image of A is A where k is the perpen- dicular bisector of AAr. 2. If point A is on k, the image of A is A. • A figure has line symmetry when the figure is its own image under a line reflection. |
• A point reflection in P is a transformation plane such that: 1. If point A is not point P, then the image of A is A where P the mid- point of AAr. 2. The point P is its own image. • A translation is a transformation in a plane that moves every point in the plane the same distance in the same direction. • A translation of a units in the horizontal direction and b units in the vertical direction is a transformation in a plane such that the image of P(x, y) is P(x a, y b). 14365C06.pgs 8/2/07 5:45 PM Page 256 256 Transformations and the Coordinate Plane • A rotation is a transformation of a plane about a fixed point P through an angle of d degrees such that: 1. For A, a point that is not the fixed point, if the image of A is A, then PA PA and mAPA d. 2. The image of the center of rotation P is P. • A quarter turn is a counterclockwise rotation of 90°. • A composition of transformations is a combination of two transformations in which the first transformation produces an image and the second transformation is performed on that image. • A glide reflection is a composition of transformations of the plane that consists of a line reflection and a translation in the direction of the line of reflection in either order. • An isometry is a transformation that preserves distance. • A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr are the same ray and OP = kOP. 3. When k is negative and the image of P is P, then opposite rays and OP = kOP. h OP and h OPr are • A function is a set of ordered pairs in which no two pairs have the same first element. • The set of first elements is the domain of the function. • The set of second elements is the range of the function. • A direct isometry is a transformation that preserves distance and orienta- tion. • An opposite isometry is a transformation that preserves distance and reverses orientation. Postulates 6.1 Two points are on the same horizontal line if and only if they have the same y-coordinates. 6.2 The length of a horizontal line segment is the absolute value of the |
differ- ence of the x-coordinates. 6.3 Two points are on the same vertical line if and only if they have the same x-coordinate. 6.4 The length of a vertical line segment is the absolute value of the difference of the y-coordinates. 6.5 Each vertical line is perpendicular to each horizontal line. Theorems and Corollaries 6.1 Under a line reflection, distance is preserved. 6.1a Under a reflection, angle measure is preserved. 6.1b Under a reflection, collinearity is preserved. 6.1c Under a reflection, midpoint is preserved. 14365C06.pgs 7/12/07 2:57 PM Page 257 Review Exercises 257 6.2 Under a reflection in the y-axis, the image of P(a, b) is P(a, b). 6.3 Under a reflection in the x-axis, the image of P(a, b) is P(a, b). 6.4 Under a reflection in the line y x, the image of P(a, b) is P(b, a). 6.5 Under a point reflection, distance is preserved. 6.5a Under a point reflection, angle measure is preserved. 6.5b Under a point reflection, collinearity is preserved. 6.5c Under a point reflection, midpoint is preserved. 6.6 Under a reflection in the origin, the image of P(a, b) is P(a, b). 6.7 Under a translation, distance is preserved. 6.7a Under a translation, angle measure is preserved. 6.7b Under a translation, collinearity is preserved. 6.7c Under a translation, midpoint is preserved. 6.8 Distance is preserved under a rotation about a fixed point. 6.9 Under a counterclockwise rotation of 90° about the origin, the image of P(a, b) is P(b, a). 6.10 Under a glide reflection, distance is preserved. VOCABULARY 6-1 Coordinate plane • x-axis • y-axis • Origin • Coordinates • Ordered pair • x-coordinate (abscissa) • y-coordinate (ordinate) • (x, y) 6-2 Line of reflection • Line reflection • Fixed point • Transformation • Preimage • Image • Reflection in line k • rk |
• Axis of symmetry • Line symmetry 6-3 ry-axis • rx-axis • ryx 6-4 Point reflection in P • RP • Point symmetry • RO 6-5 Translation • Translation of a units in the horizontal direction and b units in the vertical direction • Ta,b • Translational symmetry 6-6 Rotation • RP,d • Positive rotation • Negative rotation • Rotational symmetry • Quarter turn 6-7 Composition of transformations • Glide reflection • Isometry 6-8 Dilation • Dk 6-9 Function • Domain • Range • Orientation • Direct isometry • Opposite Isometry REVIEW EXERCISES 1. Write the equation of the vertical line that is 2 units to the left of the origin. 2. Write the equation of the horizontal line that is 4 units below the origin. 14365C06.pgs 7/12/07 2:57 PM Page 258 258 Transformations and the Coordinate Plane 3. a. On graph paper, draw the polygon ABCD whose vertices are A(4, 0), B(0, 0), C(3, 3), and D(4, 3). b. Find the area of polygon ABCD. In 4–11: a. Find the image of P(5, 3) under each of the given transformations. b. Name a fixed point of the transformation if one exists. 4. rx-axis 8. rx-axis 7. T0,3 11. R(2,2) 12. Draw a quadrilateral that has exactly one line of symmetry. 6. R90° 10. ry 5 x 5. RO 9. + rx-axis + ry 5 x + T2,2 rx-axis + T5,23 13. Draw a quadrilateral that has exactly four lines of symmetry. 14. Print a letter that has rotational symmetry. 15. The letters S and N have point symmetry. Print another letter that has point symmetry. 16. What transformations are opposite isometries? 17. What transformation is not an isometry? 18. a. On graph paper, locate the points A(3, 2), B(3, 7), and C(2, 7). Draw ABC. b. Draw ABC, the image of ABC under a reflection in the origin, and write the coordinates of its vertices. c. Draw ABC, the image of ABC under a reflection in the y-axis, |
and write the coordinates of its vertices. d. Under what single transformation is ABC the image of ABC? 19. a. On graph paper, locate the points R(4, 1), S(1, 1), and T(1, 2). Draw RST. b. Draw RST, the image of RST under a reflection in the origin, and write the coordinates of its vertices. c. Draw RST, the image of RST under a reflection in the line whose equation is y 4, and write the coordinates of its vertices. 20. a. Write the coordinates of the image, under the correspondence (x, y) → (x, 0), of each of the following ordered pairs: (3, 5), (3, 3), (1, 1), (1, 5). b. Explain why (x, y) → ( x, 0) is not a transformation. 21. The vertices of MAT have coordinates M(1, 3), A(2, 2), and T(2, 2). + D3. + rx-axis equivalent transformations? Justify your a. Find MAT, the image of MAT under the composition b. Find MAT, the image of MAT under the composition c. Are rx-axis D3 and. D3 + rx-axis + D3 rx-axis answer. 14365C06.pgs 8/2/07 5:45 PM Page 259 Cumulative Review 259 Exploration Designs for wallpaper, wrapping paper, or fabric often repeat the same pattern in different arrangements. Such designs are called tessellations. Find examples of the use of line reflections, point reflections, translations and glide reflections in these designs. CUMULATIVE REVIEW Chapters 1–6 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The product 7a(3a 1) can be written as (1) –21a 7a (2) –21a 7a (3) –21a2 7a (4) –21a2 1 2. In the coordinate plane, the point whose coordinates are (2, 0) is (1) on the x-axis (2) on the y-axis 3. If D is not on ABC (1) AD DC AC (2) AD DC AC, then (3) in the first quadrant (4) in the fourth quadrant (3) AD DC |
AC (4) AB BC AD DC 4. If is the perpendicular bisector of DBE, which of the following could g ABC be false? (1) AD AE (4) AB BC 5. DEF is not congruent to LMN, DE LM, and EF MN. Which of (3) CD CE (2) DB BE the following must be true? (1) DEF and LMN are not both right triangles. (2) mD mL (3) mF mN (4) mE mM 6. Under a reflection in the origin, the image of (2, 4) is (1) (2, 4) (2) (2, 4) (3) (2, 4) (4) (4, 2) 7. If PQ RQ, then which of the following must be true?. PR (1) Q is the midpoint of (2) Q is on the perpendicular bisector of (3) PQ QR PR (4) Q is between P and R. PR. 14365C06.pgs 7/12/07 2:57 PM Page 260 260 Transformations and the Coordinate Plane 8. Which of the following always has a line of symmetry? (1) a right triangle (2) a scalene triangle (3) an isosceles triangle (4) an acute triangle 9. In the coordinate plane, two points lie on the same vertical line. Which of the following must be true? (1) The points have the same x-coordinates. (2) The points have the same y-coordinates. (3) The points lie on the y-axis. (4) The points lie on the x-axis. 10. Angle A and angle B are complementary angles. If mA x 42 and mB 2x 12, the measure of the smaller angle is (1) 20 (2) 28 (3) 62 (4) 88 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The image of ABC under a reflection in the line y x is ADE. If the coordinates of A are (2, b), what is the value of b? Explain your answer. 12. What are the coordinates of the mid |
point of a line segment whose end- points are A(2, 5) and B(2, 3)? Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. In quadrilateral ABCD, prove that ABC ADC if. BED pendicular bisector of is the per- AEC 14. The measure of R is 12 degrees more than three times the measure of S. If S and R are supplementary angles, find the measure of each angle. 14365C06.pgs 7/12/07 2:57 PM Page 261 Cumulative Review 261 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The measures of the angles of ABC are unequal (mA mB, mB mC, and mA mC). Prove that ABC is a scalene triangle. 16. Given: T3,2(ABC) ABC and XYZ with XY AB, YZ BC, and Y B. Prove: ABC XYZ 14365C07.pgs 7/10/07 8:45 AM Page 262 CHAPTER 7 CHAPTER TABLE OF CONTENTS 7-1 Basic Inequality Postulates 7-2 Inequality Postulates Involving Addition and Subtraction 7-3 Inequality Postulates Involving Multiplication and Division 7-4 An Inequality Involving the Lengths of the Sides of a Triangle 7-5 An Inequality Involving an Exterior Angle of a Triangle 7-6 Inequalities Involving Sides and Angles of a Triangle Chapter Summary Vocabulary Review Exercises Cumulative Review 262 GEOMETRIC INEQUALITIES Euclid’s Proposition 20 of Book 1 of the Elements states,“In any triangle, two sides taken together in any manner are greater than the remaining one.” The Epicureans, a group of early Greek philosophers, ridiculed this theorem, stating that it is evident even to a donkey since if food is placed at one vertex of a triangle and the donkey at another, the donkey will make his way |
along one side of the triangle rather than traverse the other two, to get to the food. But no matter how evident the truth of a statement may be, it is important that it be logically established in order that it may be used in the proof of theorems that follow. Many of the inequality theorems of this chapter depend on this statement for their proof. 14365C07.pgs 7/10/07 8:45 AM Page 263 7-1 BASIC INEQUALITY POSTULATES Basic Inequality Postulates 263 Each time the athletes of the world assemble for the Olympic Games, they attempt to not only perform better than their competitors at the games but also to surpass previous records in their sport. News commentators are constantly comparing the winning time of a bobsled run or a 500-meter skate with the world records and with individual competitors’ records. In previous chapters, we have studied pairs of congruent lines and pairs of congruent angles that have equal measures. But pairs of lines and pairs of angles are often not congruent and have unequal measures. In this chapter, we will apply the basic inequality principles that we used in algebra to the lengths of line segments and the measures of angles. These inequalities will enable us to formulate many important geometric relationships. Postulate Relating a Whole Quantity and Its Parts In Chapter 3 we stated postulates of equality. Many of these postulates suggest related postulates of inequality. Consider the partition postulate: A whole is equal to the sum of all its parts. This corresponds to the following postulate of inequality: Postulate 7.1 A whole is greater than any of its parts. In arithmetic: Since 14 9 5, then 14 9 and 14 5. In algebra: If a, b, and c represent positive numbers and a b c, then a b and a c. In geometry: The lengths of line segments and the measures of angles are positive numbers. Consider these two applications: A E C B D F G ACB is a line segment, then AB AC CB, AB AC, and AB CB. • If • If DEF and FEG are adjacent angles, mDEG mDEF mFEG, mDEG mDEF, and mDEG mFEG. Transitive Property Consider this statement of the transitive property of equality: If a, b, and c are real numbers such that a b and b c, then a c. 14365C07.pgs 7/10/07 8 |
:45 AM Page 264 264 Geometric Inequalities This corresponds to the following transitive property of inequality: Postulate 7.2 If a, b, and c are real numbers such that a b and b c, then a c. In arithmetic: If 12 7 and 7 3, then 12 3. In algebra: If 5x 1 2x and 2x 16, then 5x 1 16. In geometry: If BA BD and BD BC, then BA BC. Also, if mBCA mBCD and mBCD mBAC, then mBCA mBAC. C B D A Substitution Postulate Consider the substitution postulate as it relates to equality: A quantity may be substituted for its equal in any statement of equality. Substitution also holds for inequality, as demonstrated in the following postulate: Postulate 7.3 A quantity may be substituted for its equal in any statement of inequality. In arithmetic: If 10 2 5 and 2 5 7, then 10 7. In algebra: If 5x 1 2y and y 4, then 5x 1 2(4). In geometry: If AB BC and BC AC, then AB AC. Also, if mC mA and mA = mB, then mC mB. C A B The Trichotomy Postulate We know that if x represents the coordinate of a point on the number line, then x can be a point to the left of 3 when x 3, x can be the point whose coordinate is 3 if x 3, or x can be a point to the right of 3 if x 3. We can state this as a postulate that we call the trichotomy postulate, meaning that it is divided into three cases. Postulate 7.4 Given any two quantities, a and b, one and only one of the following is true: a b a b. a b or or 14365C07.pgs 7/10/07 8:45 AM Page 265 Basic Inequality Postulates 265 EXAMPLE 1 Given: mDAC mDAB mBAC and mDAB mABC A D Prove: mDAC mABC C B Proof Statements Reasons 1. mDAC mDAB mBAC 2. mDAC mDAB 1. Given. 2. A whole is greater than any of its 3. mDAB mABC 4. mDAC mABC parts. 3. Given. 4. Transitive property of inequality. EXAM |
PLE 2 Given: Q is the midpoint of PS and RS QS. Prove: RS PQ P Q R S Proof Statements Reasons 1. Q is the midpoint of. PS 1. Given. 2. PQ > QS 3. PQ = QS. 4. RS QS 5. RS PQ 2. The midpoint of a line segment is the point that divides the segment into two congruent segments. 3. Congruent segments have equal measures. 4. Given. 5. Substitution postulate. 14365C07.pgs 7/10/07 8:45 AM Page 266 266 Geometric Inequalities Exercises Writing About Mathematics 1. Is inequality an equivalence relation? Explain why or why not. 2. Monica said that when AB BC is false, AB BC must be true. Do you agree with Monica? Explain your answer. Developing Skills In 3–12: a. Draw a diagram to illustrate the hypothesis and tell whether each conclusion is true or false. b. State a postulate or a definition that justifies your answer. 3. If ADB is a line segment, then DB AB., then CD DA CA. AC 4. If D is not on 5. If BCD DCA = BCA, then mBCD mBCA. h DB are opposite rays with point C not on h DA h DB 6. If and h DA or, then mBDC mCDA 180. h DA h DB and 7. If are opposite rays and mBDC 90, then mCDA 90. is a line segment, then DA BD, or DA BD, or DA BD. ADB 8. If 9. If AT AS and AS AR, then AT AR. 10. If m1 m2 and m2 m3, then m1 m3. 11. If SR KR and SR TR, then TR KR. 12. If m3 m2 and m2 m1, then m3 m1. Applying Skills 13. Given: ABC is isosceles, AC BC, mCBD mCBA Prove: mCBD mA C A B D 14. Given: PQRS Prove: a. PR PQ and PQ RS b. PR RS P Q R S 14365C07.pgs 7/10/07 8:45 AM Page 267 In 15 and 16, use the figure to the right. N Inequality Postulates Involving |
Addition and Subtraction 267 KLM 15. If 16. If KM KN, KN NM, and NM NL, prove that KM NL., prove that KM NM. NM and LM LK M 7-2 INEQUALITY POSTULATES INVOLVING ADDITION AND SUBTRACTION Postulates of equality and examples of inequalities involving the numbers of arithmetic can help us to understand the inequality postulates presented here. Consider the addition postulate: If equal quantities are added to equal quantities, then the sums are equal. Addition of inequalities requires two cases: Postulate 7.5 If equal quantities are added to unequal quantities, then the sums are unequal in the same order. Postulate 7.6 If unequal quantities are added to unequal quantities in the same order, then the sums are unequal in the same order. E D C In arithmetic: Since 12 5, then 12 3 5 3 or 15 8. Since 12 5 and 3 2, then 12 3 5 2 or 15 7. In algebra: If x 5 10, then x 5 5 10 5 or x 15. If x 5 10 and 5 3, then x 5 5 10 3 or x 13. In geometry: If and AB CD, then AB BC BC CD or ABCD AC BD. If ABCDE AB BC CD DE or AC CE., AB CD, and BC DE, then B A We can subtract equal quantities from unequal quantities without changing the order of the inequality, but the result is uncertain when we subtract unequal quantities from unequal quantities. Consider the subtraction postulate: If equal quantities are subtracted from equal quantities, then the differences are equal. 14365C07.pgs 7/10/07 8:45 AM Page 268 268 Geometric Inequalities Subtraction of inequalities is restricted to a single case: Postulate 7.7 If equal quantities are subtracted from unequal quantities, then the differences are unequal in the same order. However, when unequal quantities are subtracted from unequal quantities, the results may or may not be unequal and the order of the inequality may or may not be the same. For example: • 5 2 and 4 1, but it is not true that 5 4 2 1 since 1 1. • 12 10 and 7 1, but it is not true that 12 7 10 1 since 5 9. • 12 10 and 2 1, and it is true that 12 2 10 1 since 10 9. EXAMPLE 1 Given: mBDE mCDA Prove: mB |
DC mEDA E C Proof Statements 1. mBDE mCDA 2. mBDE mEDC mEDC mCDA B D A Reasons 1. Given. 2. If equal quantities are added to unequal quantities, then the sums are unequal in the same order. 3. mBDC mBDE mEDC 3. The whole is equal to the sum of its parts. 4. mEDA mEDC mCDA 4. The whole is equal to the sum of 5. mBDC mEDA its parts. 5. Substitution postulate for inequalities. 14365C07.pgs 7/10/07 8:45 AM Page 269 Inequality Postulates Involving Addition and Subtraction 269 Exercises Writing About Mathematics 1. Dana said that 13 11 and 8 3. Therefore, 13 8 11 3 tells us that if unequal quantities are subtracted from unequal quantities, the difference is unequal in the opposite order. Do you agree with Dana? Explain why or why not. 2. Ella said that if unequal quantities are subtracted from equal quantities, then the differences are unequal in the opposite order. Do you agree with Ella? Explain why or why not. Developing Skills In 3–10, in each case use an inequality postulate to prove the conclusion. 3. If 10 7, then 18 15. 5. If x 3 12, then x 9. 7. If 8 6 and 5 3, then 13 9. 9. If y 8, then y 1 7. 4. If 4 14, then 15 25. 6. If y 5 5, then y 10. 8. If 7 12, then 5 10. 10. If a b, then 180 a 90 b. Applying Skills 11. Given: AB AD, BC DE 12. Given: AE BD, AF BF Prove: AC AE A D B C E 13. Given: mDAC mDBC and AE BE Prove: a. mEAB mEBA b. mDAB mCBA Prove: FE FD 14. In August, Blake weighed more than Caleb. In the next two months, Blake and Caleb had each gained the same number of pounds. Does Blake still weigh more than Caleb? Justify your answer. 15. In December, Blake weighed more than Andre. In the next two months, Blake lost more than Andre lost. Does Blake still weigh more than Andre? Justify your answer. 14541C07 |
.pgs 1/25/08 3:50 PM Page 270 270 Geometric Inequalities 7-3 INEQUALITY POSTULATES INVOLVING MULTIPLICATION AND DIVISION Since there are equality postulates for multiplication and division similar to those of addition and subtraction, we would expect that there are inequality postulates for multiplication and division similar to those of addition and subtraction. Consider these examples that use both positive and negative numbers. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 3 15. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 3 15. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 1 15. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 3 15. Notice that in the top four examples, we are multiplying by positive numbers and the order of the inequality does not change. In the bottom four examples, we are multiplying by negative numbers and the order of the inequality does change. These examples suggest the following postulates of inequality: Postulate 7.8 If unequal quantities are multiplied by positive equal quantities, then the products are unequal in the same order. Postulate 7.9 If unequal quantities are multiplied by negative equal quantities, then the products are unequal in the opposite order. Since we know that division by a 0 is the same as multiplication by and that a and are always either both positive or both negative, we can write similar postulates for division of inequalities. 1 a 1 a Postulate 7.10 If unequal quantities are divided by positive equal quantities, then the quotients are unequal in the same order. 14365C07.pgs 7/10/07 8:45 AM Page 271 Inequality Postulates Involving Multiplication and Division 271 Postulate 7.11 If unequal quantities are divided by negative equal quantities, then the quotients are unequal in the opposite order. Care must be taken when using inequality postulates involving multiplication and division because multiplying or dividing by a negative number will reverse the order of the inequality. EXAMPLE 1 Given: BA 3BD, BC 3BE, and BE BD Prove |
: BC BA B D E A C Proof Statements Reasons 1. BE BD 2. 3BE 3BD 3. BC 3BE, BA 3BD 4. BC BA 1. Given. 2. If unequal quantities are multi- plied by positive equal quantities, then the products are unequal in the same order. 3. Given. 4. Substitution postulate for inequalities. EXAMPLE 2 Given: mABC mDEF, ABC, h EH h BG bisects DEF. bisects C G Prove: mABG mDEH B A E F H D Proof An angle bisector separates the angle into two congruent parts. Therefore, the measure of each part is one-half the measure of the angle that was bisected, so 1 mABG 2m/DEF and mDEH 1 2m/ABC. Since we are given that mABC mDEF, 1 2m/DEF because if unequal quantities are multiplied by positive equal quantities, the products are unequal in the same order. Therefore, by the substitution postulate for inequality, mABG mDEH. 1 2m/ABC 14365C07.pgs 7/10/07 8:45 AM Page 272 272 Geometric Inequalities Exercises Writing About Mathematics 1. Since 1 2, is it always true that a 2a? Explain why or why not. 2. Is it always true that if a b and c d, then ac bd? Justify your answer. Developing Skills In 3–8, in each case state an inequality postulate to prove the conclusion. 3. If 8 7, then 24 21. 5. If 8 6, then 4 3. 7. If 4, then x 8. x 2 4. If 30 35, then 6 7. 6. If 3x 15, then x 5. 8. If 3, then y 18. y 6 In 9–17: If a, b, and c are positive real numbers such that a b and b c, tell whether each relationship is always true, sometimes true, or never true. If the statement is always true, state the postulate illustrated. If the statement is sometimes true, give one example for which it is true and one for which it is false. If the statement is never true, give one example for which it is false. 9. ac bc 12. a c b c 15. c. b a c Applying Skills 10. a c b c 13. a b |
b c 16. ac bc 11. c a c b c a. c 14. b 17. a c 18. Given: BD BE, D is the midpoint BA of Prove: BA BC, E is the midpoint of. BC B D A E C 20. Given: AB AD, AE AF 1 2AD Prove: AE AF 1 2AB, D F C A E B 19. Given: mDBA mCAB, mCBA 2mDBA, mDAB 2mCAB Prove: mCBA mDAB D C E A B 21. Given: mCAB mCBA, bisects CAB, BE AD bisects CBA. Prove: mDAB mEBA C D B E A 14365C07.pgs 7/10/07 8:45 AM Page 273 An Inequality Involving the Lengths of the Sides of a Triangle 273 7-4 AN INEQUALITY INVOLVING THE LENGTHS OF THE SIDES OF A TRIANGLE The two quantities to be compared are often the lengths of line segments or the distances between two points. The following postulate was stated in Chapter 4. The shortest distance between two points is the length of the line segment joining these two points. The vertices of a triangle are three is noncollinear points. The length of AB, the shortest distance from A to B. Therefore, AB AC CB. Similarly, BC BA AC and AC AB BC. We have just proved the following theorem, called the triangle inequality theorem: AB C A B Theorem 7.1 The length of one side of a triangle is less than the sum of the lengths of the other two sides. In the triangle shown above, AB AC BC. To show that the lengths of three line segments can be the measures of the sides of a triangle, we must show that the length of any side is less than the sum of the other two lengths of the other two sides. EXAMPLE 1 Which of the following may be the lengths of the sides of a triangle? (1) 4, 6, 10 (2) 8, 8, 16 (3) 6, 8, 16 (4) 10, 12, 14 Solution The length of a side of a triangle must be less than the sum of the lengths of the other two sides. If the lengths of the sides are a b c, then a b means that a b c and b c |
means that b c a. Therefore, we need only test the longest side. (1) Is 10 4 6? No (2) Is 16 8 8? No (3) Is 16 6 8? No (4) Is 14 10 12? Yes Answer 14365C07.pgs 7/10/07 8:45 AM Page 274 274 Geometric Inequalities EXAMPLE 2 Two sides of a triangle have lengths 3 and 7. Find the range of possible lengths of the third side. Solution (1) Let s length of third side of triangle. (2) Of the lengths 3, 7, and s, the longest side is either 7 or s. (3) If the length of the longest side is s, then s 3 7 or s 10. (4) If the length of the longest side is 7, then 7 s 3 or 4 s. Answer 4 s 10 EXAMPLE 3 Given: Isosceles triangle ABC with AB CB and M. AC the midpoint of B Prove: AM AB A M C Proof Statements Reasons 1. AC AB CB 1. The length of one side of a triangle is less than the sum of the lengths of the other two sides. 2. AB CB 3. AC AB AB or AC 2AB 2. Given. 3. Substitution postulate for. AC 4. M is the midpoint of 5. AM MC 6. AC AM MC 7. AC AM AM 2AM 8. 2AM 2AB inequalities. 4. Given. 5. Definition of a midpoint. 6. Partition postulate. 7. Substitution postulate. 8. Substitution postulate for inequality. 9. AM AB 9. Division postulate for inequality. 14365C07.pgs 7/10/07 8:45 AM Page 275 An Inequality Involving the Lengths of the Sides of a Triangle 275 Exercises Writing About Mathematics 1. If 7, 12, and s are the lengths of three sides of a triangle, and s is not the longest side, what are the possible values of s? 2. a. If a b c are any real numbers, is a b c always true? Justify your answer. b. If a b c are the lengths of the sides of a triangle, is a b c always true? Justify your answer. Developing Skills In 3–10, tell in each case whether the given lengths can be the measures of the sides of a triangle. 3. 3 |
, 4, 5 7. 2, 2, 3 4. 5, 8, 13 8. 1, 1, 2 5. 6, 7, 10 9. 3, 4, 4 6. 3, 9, 15 10. 5, 8, 11 In 11–14, find values for r and t such that the inequality r s t best describes s, the length of the third sides of a triangle for which the lengths of the other two sides are given. 11. 2 and 4 12. 12 and 31 13. 13 2 and 13 2 14. 9.6 and 12.5 15. Explain why x, 2x, and 3x cannot represent the lengths of the sides of a triangle. 16. For what values of a can a, a 2, a 2 represent the lengths of the sides of a triangle? Justify your answer. Applying Skills 17. Given: ABCD is a quadrilateral. Prove: AD AB BC CD D C A B 19. Given: Point P in the interior of XYZ, YPQ Prove: PY PZ XY XZ 18. Given: ABC with D a point on and AD DC. BC Prove: AB BC B D C Q A X P Y Z 14365C07.pgs 7/10/07 8:45 AM Page 276 276 Geometric Inequalities Hands-On Activity One side of a triangle has a length of 6. The lengths of the other two sides are integers that are less than or equal to 6. a. Cut one straw 6 inches long and two sets of straws to integral lengths of 1 inch to 6 inches. Determine which lengths can represent the sides of a triangle. Use geometry software to determine which lengths can represent the sides of a triangle. b. List all sets of three integers that can be the lengths of the sides of the triangle. Or For example, is one set of lengths. {6, 3, 5} c. List all sets of three integers less than or equal to 6 that cannot be the lengths of the sides of the triangle. d. What patterns emerge in the results of parts b and c? 7-5 AN INEQUALITY INVOLVING AN EXTERIOR ANGLE OF A TRIANGLE Exterior Angles of a Polygon At each vertex of a polygon, an angle is formed that is the union of two sides of the polygon. Thus, for polygon ABCD, DAB is an angle of |
the polygon, often called an interior angle. If, at h vertex A, we draw, AD we form BAE, an exterior angle of the polygon at vertex A., the opposite ray of h AE C D A E B DEFINITION An exterior angle of a polygon is an angle that forms a linear pair with one of the interior angles of the polygon. 14365C07.pgs 7/10/07 8:45 AM Page 277 An Inequality Involving an Exterior Angle of a Triangle 277 h AF h AB At vertex A, we can also draw, the, to form DAF, another opposite ray of exterior angle of the polygon at vertex A. At each vertex of a polygon, two exterior angles can be drawn. Each of these exterior angles forms a linear pair with the interior angle at A, and the angles in a linear pair are supplementary. The two exterior angles at A are congruent angles because they are vertical angles. Either can be drawn as the exterior angle at A. C D F A E B Exterior Angles of a Triangle An exterior angle of a triangle is formed outside the triangle by extending a side of the triangle. The figure to the left shows ABC whose three interior angles are CAB, ABC, and BCA. By extending each side of ABC, three exterior angles are formed, namely, DAC, EBA, and FCB. For each exterior angle, there is an adjacent interior angle and two remote or nonadjacent interior angles. For ABC, these angles are as follows: FC B E D A Vertex Exterior Angle Adjacent Interior Angle Nonadjacent Interior Angles A B C DAC EBA FCB CAB ABC BCA ABC and BCA CAB and BCA CAB and ABC With these facts in mind, we are now ready to prove another theorem about inequalities in geometry called the exterior angle inequality theorem. Theorem 7.2 The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. Given ABC with exterior BCD at vertex C; A and B are nonadjacent interior angles with respect to BCD. Prove mBCD mB B E M A C D 14541C07.pgs 1/25/08 3:50 PM Page 278 278 Geometric Inequalities Proof E B M Statements Reasons 1. Let M be the midpoint of BC. 1. Every line segment has one and 2. Draw h |
AM, extending the ray A C D through M to point E so only one midpoint. 2. Two points determine a line. A line segment can be extended that AM. EM to any length.. EC 3. Draw 4. mBCD mBCE mECD 3. Two points determine a line. 4. A whole is equal to the sum of its 5. BM CM EM AM 6. 7. AMB EMC 8. AMB EMC 9. B MCE parts. 5. Definition of midpoint. 6. Construction (step 2). 7. Vertical angles are congruent. 8. SAS (steps 5, 7, 6). 9. Corresponding parts of congruent triangles are congruent. 10. mBCD mMCE 10. A whole is greater than any of its 11. mBCD mB parts. 11. Substitution postulate for inequalities. These steps prove that the measure of an exterior angle is greater than the measure of one of the nonadjacent interior angles, B. A similar proof can be used to prove that the measure of an exterior angle is greater than the measure of the other nonadjacent interior angle, A. This second proof uses N, the midpoint of, and a point F extending ray h BC AC through C. The details of this proof will be left to the student. (See exercise 14.), a line segment BN > NG BNG with EXAMPLE 1 The point D is on AB of ABC. a. Name the exterior angle at D of ADC. A D b. Name two nonadjacent interior angles of the exterior angle at D of ADC. C B c. Why is mCDB mDCA? d. Why is AB AD? 14365C07.pgs 7/10/07 8:45 AM Page 279 An Inequality Involving an Exterior Angle of a Triangle 279 Solution a. CDB b. DCA and A c. The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. d. The whole is greater than any of its parts. EXAMPLE 2 Given: Right triangle ABC, mC 90, BAD is an exterior angle at A. C Prove: BAD is obtuse. A D B Proof Statements Reasons 1. BAD is an exterior angle. 2. mBAD mC 1. Given. 2. Exterior angle inequality 3. mC 90 4. mBAD 90 |
5. mBAD mBAC 180 6. 180 mBAD 7. 180 mBAD 90 8. BAD is obtuse. theorem. 3. Given. 4. Substitution postulate for inequalities. 5. If two angles form a linear pair, then they are supplementary. 6. The whole is greater than any of its parts. 7. Steps 4 and 6. 8. An obtuse angle is an angle whose degree measure is greater than 90 and less than 180. 14541C07.pgs 1/25/08 3:50 PM Page 280 280 Geometric Inequalities Exercises Writing About Mathematics 1. Evan said that every right triangle has at least one exterior angle that is obtuse. Do you agree with Evan? Justify your answer. 2. Connor said that every right triangle has at least one exterior angle that is a right angle. Do you agree with Connor? Justify your answer. Developing Skills T 3. a. Name the exterior angle at R. b. Name two nonadjacent interior angles of the exterior angle at R. S R P In 4–13, ABC is scalene and side AB. CM is a median to a. Tell whether each given statement is True or False. b. If the statement is true, state the definition, postulate, or A M B C theorem that justifies your answer. 4. AM MB 6. mAMC mABC 8. mCMB mACM 10. BA MB 12. mBCA mMCA Applying Skills 14. Given: ABC with exterior BCD at vertex C; A and B are nonadjacent interior angles with respect to BCD. Prove: mBCD mA 5. mACB mACM 7. AB AM 9. mCMB mCAB 11. mACM mBCM 13. mBMC mAMC 15. Given: ABD DBE ABE and ABE EBC ABC Prove: mABD mABC (Complete the proof of Theorem 7.2). A B A C D B D E C 14365C07.pgs 7/10/07 8:45 AM Page 281 Inequalities Involving Sides and Angles of a Triangle 281 16. Given: Isosceles DEF with DE FE and exterior EFG Prove: mEFG mEFD E 17. Given: Right ABC with mC 90 Prove: A is acute. B D F G 18. |
Given: SMR with STM extended through M to P Prove: mRMP mSRT R A C 19. Given: Point F not on › ‹ ABCDE and FC FD Prove: mABF mEDF F S T M P A CB D E 7-6 INEQUALITIES INVOLVING SIDES AND ANGLES OF A TRIANGLE A 22° We know that if the lengths of two sides of a triangle are equal, then the measures of the angles opposite these sides are equal. Now we want to compare the measures of two angles opposite sides of unequal length. Let the measures of the sides of ABC be AB 12, BC 5, and CA 9. 12 9 115° C 43° 5 B Theorem 7.3 Write the lengths in order: 12 9 5 Name the sides in order: AB CA BC Name the angles opposite these sides in order: mC mB mA Notice how the vertex of the angle opposite a side of the triangle is always the point that is not an endpoint of that side. If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. 14365C07.pgs 7/10/07 8:45 AM Page 282 282 Geometric Inequalities To prove this theorem, we will extend the shorter side of a triangle to a length equal to that of the longer side, forming an isosceles triangle. We can then use the isosceles triangle theorem and the exterior angle inequality theorem to compare angle measures. Given ABC with AB BC Prove mACB mBAC A B C D Proof Statements 1. ABC with AB BC. Reasons 1. Given. 2. Extend through C to point D so that BD BA. BC. 3. Draw AD 4. ABD is isosceles. 5. mBAD mBDA 6. For ACD, mBCA mBDA. 7. mBCA mBAD 2. A line segment may be extended to any length. 3. Two points determine a line. 4. Definition of isosceles triangle. 5. Base angles of an isosceles triangle are equal in measure. 6. Exterior angle inequality theorem. 7. Substitution postulate for inequalities. 8. mBAD mBAC 8. A whole is greater than any of its parts. 9. mBCA mBAC |
9. Transitive property of inequality. The converse of this theorem is also true, as can be seen in this example: Let the measures of the angles of ABC be mA 40, mB 80, and mC = 60. Write the angle measures in order: 80 60 40 Name the angles in order: mB mC mA Name the sides opposite these angles in order: AC AB BC If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. A 3.7 40° 4.2 B 2.8 80° 60° C Theorem 7.4 14365C07.pgs 7/10/07 8:45 AM Page 283 Inequalities Involving Sides and Angles of a Triangle 283 We will write an indirect proof of this theorem. Recall that in an indirect proof, we assume the opposite of what is to be proved and show that the assumption leads to a contradiction. Given DEF with mD mE F Prove FE FD D E Proof By the trichotomy postulate: FE FD or FE FD or FE FD. We assume the negation of the conclusion, that is, we assume FE FD. Therefore, either FE FD or FE FD. If FE FD, then mD mE because base angles of an isosceles triangle are equal in measure. This contradicts the given premise, mD mE. Thus, FE FD is a false assumption. If FE FD, then, by Theorem 7.3, we must conclude that mD mE. This also contradicts the given premise that mD mE. Thus, FE FD is also a false assumption. Since FE FD and FE FD are both false, FE FD must be true and the theorem has been proved. EXAMPLE 1 One side of ABC is extended to D. If mA 45, mB 50, and mBCD 95, which is the longest side of ABC? B Solution The exterior angle and the interior angle at vertex C form a linear pair and are supplementary. Therefore: mBCA 180 mBCD 180 95 85 A C D Since 85 50 45, the longest side of the triangle is BCA. Answer BA, the side opposite EXAMPLE 2 In ADC, CB that CD CA. is drawn to ABD and CA > CB. Prove C Proof Consider CBD. The measure of an exterior angle is greater than the measure of a non |
adjacent interior angle, so mCBA mCDA. Since angles of an isosceles triangle have equal measures, so mA mCBA. A quantity may be substituted for its equal in an inequality, so mA mCDA. A, ABC is isosceles. The base CA > CB D B 14365C07.pgs 7/10/07 8:45 AM Page 284 284 Geometric Inequalities C If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side is opposite the larger angle. Therefore, CD AC. A B D Exercises Writing About Mathematics 1. a. Write the contrapositive of the statement “If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal.” b. Is this contrapositive statement true? 2. The Isosceles Triangle Theorem states that if two sides of a triangle are congruent, then the angles opposite these sides are congruent. a. Write the converse of the Isosceles Triangle Theorem. b. How is this converse statement related to the contrapositive statement written in exer- cise 1? Developing Skills 3. If AB 10, BC 9, and CA 11, name the largest angle of ABC. 4. If mD 60, mE 70, and mF 50, name the longest side of DEF. In 5 and 6, name the shortest side of ABC, using the given information. 5. In ABC, mC 90, mB 35, and mA 55. 6. In ABC, mA 74, mB 58, and mC 48. In 7 and 8, name the smallest angle of ABC, using the given information. 7. In ABC, AB 7, BC 9, and AC 5. 8. In ABC, AB 5, BC 12, and AC 13. 9. In RST, an exterior angle at R measures 80 degrees. If mS mT, name the shortest side of the triangle. 10. If ABD is an exterior angle of BCD, mABD 118, mD 60, and mC 58, list the sides of BCD in order starting with the longest. 11. If EFH is an exterior angle of FGH, mEFH 125, mG 65, mH 60, list the sides of FGH in order |
starting with the shortest. 12. In RST, S is obtuse and mR mT. List the lengths of the sides of the triangle in order starting with the largest. 14365C07.pgs 8/2/07 5:46 PM Page 285 Applying Skills 13. Given: C is a point that is not on, ABD mABC mCBD. Prove: AC BC Chapter Summary 285 C A B D 14. Let ABC be any right triangle with the right angle at C and hypotenuse. AB a. Prove that A and B are acute angles. b. Prove that the hypotenuse is the longest side of the right triangle. 15. Prove that every obtuse triangle has two acute angles. CHAPTER SUMMARY Definitions to Know • An exterior angle of a polygon is an angle that forms a linear pair with one of the interior angles of the polygon. • Each exterior angle of a triangle has an adjacent interior angle and two remote or nonadjacent interior angles. Postulates 7.1 A whole is greater than any of its parts. If a, b, and c are real numbers such that a b and b c, then a c. 7.2 7.3 A quantity may be substituted for its equal in any statement of inequality. 7.4 Given any two quantities, a and b, one and only one of the following is 7.5 7.6 7.7 7.8 7.9 true: a b, or a b, or a b. If equal quantities are added to unequal quantities, then the sums are unequal in the same order. If unequal quantities are added to unequal quantities in the same order, then the sums are unequal in the same order. If equal quantities are subtracted from unequal quantities, then the differences are unequal in the same order. If unequal quantities are multiplied by positive equal quantities, then the products are unequal in the same order. If unequal quantities are multiplied by negative equal quantities, then the products are unequal in the opposite order. 7.10 If unequal quantities are divided by positive equal quantities, then the quotients are unequal in the same order. 7.11 If unequal quantities are divided by negative equal quantities, then the quotients are unequal in the opposite order. Theorems 7.1 The length of one side of a triangle is less than the sum of the lengths of the other two sides. 7.2 The measure of an exterior angle of a triangle is greater |
than the measure of either nonadjacent interior angle. 14365C07.pgs 7/10/07 8:45 AM Page 286 286 Geometric Inequalities 7.3 If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. 7.4 If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. VOCABULARY 7-1 Transitive property of inequality • Trichotomy postulate 7-4 Triangle inequality theorem 7-5 Exterior angle of a polygon • Adjacent interior angle • Nonadjacent interior angle • Remote interior angle • Exterior angle inequality theorem REVIEW EXERCISES In 1–8, state a definition, postulate, or theorem that justifies each of the following statements about the triangles in the figure. D 1. AC BC 2. If DA DB and DB DC, then DA DC. A 3. mDBC mA 4. If mC mCDB, then DB BC. 5. If DA DB, then DA AC DB AC. 6. DA AC DC 7. If mA mC, then DC DA. 8. mADC mADB B C, AE BD, 9. Given:, BDC AEC and EC DC Prove: mB mA 10. Given: ABC CDA, AD DC Prove: a. mACD mCAD does not bisect A. B D b. AC B C A E C A D 14365C07.pgs 8/2/07 5:47 PM Page 287 Review Exercises 287 11. In isosceles triangle ABC, C, prove that DB DA. CA CB. If D is a point on AC between A and 12. In isosceles triangle RST, RS ST. Prove that SRP, the exterior angle at R, is congruent to STQ, the exterior angle at T. 13. Point B is 4 blocks north and 3 blocks east of A. All streets run north and south or east and west except a street that slants from C to B. Of the three paths from A and B that are marked: a. Which path is shortest? Justify your answer. b. Which path is longest? Justify your answer Exploration The Hinge Theorem states: If |
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