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two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is greater than the third side of the second triangle. 1. a. With a partner or in a small group, prove the Hinge Theorem. b. Compare your proof with the proofs of the other groups. Were different diagrams used? Were different approaches used? Were these approaches valid? The converse of the Hinge Theorem states: If the two sides of one triangle are congruent to two sides of another triangle, and the third side of the first triangle is greater than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second triangle. 2. a. With a partner or in a small group, prove the converse of the Hinge Theorem. b. Compare your proof with the proofs of the other groups. Were different diagrams used? Were different approaches used? Were these approaches valid? 14365C07.pgs 7/10/07 8:45 AM Page 288 288 Geometric Inequalities CUMULATIVE REVIEW CHAPTERS 1–7 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The solution set of the equation 2x 3.5 5x 18.2 is (1) 49 (2) 49 (3) 4.9 (4) 4.9 2. Which of the following is an example of the transitive property of inequal- ity? (1) If a b, then b a. (2) If a b, then a c b c. 3. Point M is the midpoint of (1) AM MC ABMC (2) AB MC (3) If a b and c 0, then ac bc. (4) If a b and b c, then a c.. Which of the following is not true? (3) AM BC (4) BM MC 4. The degree measure of the larger of two complementary angles is 30 more than one-half the measure of the smaller. The degree measure of the smaller is (1) 40 (4) 100 (2) 50 (3) 80 5. Which of the following could be the measures of the sides of a triangle? (1) 2, 2, 4 (2) 1, 3 |
, 5 (3) 7, 12, 20 (4) 6, 7, 12 6. Which of the following statements is true for all values of x? (1) x 5 and x 5 (2) x 5 or x 5 7. In ABC and DEF, (3) If x 5, then x 3. (4) If x 3, then x 5., and A D. In order to prove AB > DE ABC DEF using ASA, we need to prove that (1) B E BC > EF (2) C F AC > DF (3) (4) 8. Under a reflection in the y-axis, the image of (2, 5) is (2) (2, 5) (3) (2, 5) (1) (2, 5) (4) (5, 2) 9. Under an opposite isometry, the property that is changed is (1) distance (2) angle measure (3) collinearity (4) orientation 10. Points P and Q lie on the perpendicular bisector of AB. Which of the fol- is the perpendicular bisector of PQ. AB lowing statements must be true? (1) (2) PA PB and QA QB. (3) PA QA and PB QB. (4) P is the midpoint of AB or Q is the midpoint of.AB 14365C07.pgs 7/10/07 8:45 AM Page 289 Cumulative Review 289 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. Each of the following statements is true. If the snow continues to fall, our meeting will be cancelled. Our meeting is not cancelled. Can you conclude that snow does not continue to fall? List the logic principles needed to justify your answer. 12. The vertices of ABC are A(0, 3), B(4, 3), and C(3, 5). Find the coordi- nates of the vertices of ABC, the image of ABC under the composition ryx T4,5. + 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. Given: PR Prove: AP BP bisects ARB but PR is not perpendicular to ARB. 14. Given: In quadrilateral ABCD, h AC bisects DAB and h CA bisects DCB. Prove: B D 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 intersection of g PQ and g RS is T. If mPTR x, mQTS y, and mRTQ 2x y, find the measures of PTR, QTS, RTQ, and PTS. 16. In ABC, mA mB and DCB is an exterior angle at C. The mea- sure of BCA 6x 8, and the measure of DCB 4x 12. a. Find mBCA and mDCB. b. List the interior angles of the triangle in order, starting with the smallest. c. List the sides of the triangles in order starting with the smallest. 14365C08.pgs 7/10/07 8:46 AM Page 290 CHAPTER 8 CHAPTER TABLE OF CONTENTS 8-1 The Slope of a Line 8-2 The Equation of a Line 8-3 Midpoint of a Line Segment 8-4 The Slopes of Perpendicular Lines 8-5 Coordinate Proof 8-6 Concurrence of the Altitudes of a Triangle Chapter Summary Vocabulary Review Exercises Cumulative Review 290 SLOPES AND EQUATIONS OF LINES In coordinate geometry, a straight line can be characterized as a line whose slope is a constant. Do curves have slopes and if so, can they be determined? In the late 17th and early 18th centuries, two men independently developed methods to answer these and other questions about curves and the areas that they determine. The slope of a curve at a point is defined to be the slope of the tangent to that curve at that point. Descartes worked on the problem of finding the slope of a tangent to a curve by considering the slope of a tangent to a circle that intersected the curve at a given point and had its center on an axis. Gottfried Leibn |
iz (1646–1716) in Germany and Isaac Newton (1642–1727) in England each developed methods for determining the slope of a tangent to a curve at any point as well as methods for determining the area bounded by a curve or curves. Newton acknowledged the influence of the work of mathematicians and scientists who preceded him in his statement, “If I have seen further, it is by standing on the shoulders of giants.” The work of Leibniz and Newton was the basis of differential and integral calculus. 14365C08.pgs 7/10/07 8:46 AM Page 291 The Slope of a Line 291 8-1 THE SLOPE OF A LINE In the coordinate plane, horizontal and vertical lines are used as reference lines. Slant lines intersect horizontal lines at acute and obtuse angles. The ratio that measures the slant of a line in the coordinate plane is the slope of the line. Finding the Slope of a Line Through two points, one and only one line can be drawn. In the coordinate plane, if the coordinates of two points are given, it is possible to use a ratio to determine the measure of the slant of the line. This ratio is the slope of the line. Through the points A(1, 2) and B(2, 7), g is drawn. Let C(2, 2) be the point at which AB the vertical line through B intersects the horiis the zontal line through A. The slope of ratio of the change in vertical distance, BC, to the change in horizontal distance, AC. Since B and C are on the same vertical line, BC is the difference in the y-coordinates of B and C. Since A and C are on the same horizontal line, AC is the difference in the x-coordinates of A and C. AB y B(2, 7) 1 1 O 1 A(1, 2) x C(2, 2) slope of AB 5 change in vertical distance change in horizontal distance 5 BC AC 7 2 (22) 2 2 (21) 5 5 9 3 5 3 This ratio is the same for any segment of the line. Suppose we change the order of the points (1, 2) and (2, 7) in performing the computation. We then have: g AB slope of BA 5 change in vertical distance change in horizontal distance 5 CB CA (22) 2 7 (21) 2 2 5 5 29 23 |
5 3 The result of both computations is the same. When we compute the slope of a line that is determined by two points, it does not matter which point is considered as the first point and which the second. 14365C08.pgs 7/10/07 8:46 AM Page 292 292 Slopes and Equations of Lines Also, when we find the slope of a line using two points on the line, it does not matter which two points on the line we use because all segments of a line have the same slope as the line. Procedure To find the slope of a line: 1. Select any two points on the line. 2. Find the vertical change, that is, the change in y-values by subtracting the y-coordinates in any order. 3. Find the horizontal change, that is, the change in x-values, by subtracting the x-coordinates in the same order as the y-coordinates. 4. Write the ratio of the vertical change to the horizontal change. In general, the slope, m, of a line that passes through any two points A( ) and 2 x2 B(, is the ratio of the x1 difference of the y-values of these points to the difference of the corresponding x-values. ), where x1, y1 x2, y2 y B(x2, y2) A(x1, y1) x x2 y1 y y2 x1 g AB m y2 2 y1 x2 2 x1 slope of x The difference in x-values, x2 – x1, can be represented by x, read as y1, can be represented by y, “delta x.” Similarly, the difference in y-values, y2 read as “delta y.” Therefore, we write: O slope of a line m Dy Dx The slope of a line is positive if the line slants upward from left to right, negative if the line slants downward from left to right, or zero if the line is horizontal. If the line is vertical, it has no slope. Positive Slope g AB. Let The points C and D are two points on the coordinates of C be (1, 2) and the coordinates of D be (3, 3). As the values of x increase, g the values of y also increase. The graph of AB slants upward. slope of g AB The slope of is positive. g |
AB (3, 3) C(1, 2) 1 x 14365C08.pgs 7/10/07 8:46 AM Page 293 Negative Slope The Slope of a Line 293. Let the Points C and D are two points on coordinates of C be (2, 3) and the coordinates of D be (5, 1). As the values of x increase, the val- g EF ues of y decrease. The graph of ward. g EF slants down- g EF 5 1 2 3 5 2 2 5 22 3 slope of g EF The slope of is negative(2, 3) 3 D(5, 1) F x Zero Slope Points C and D are two points on. Let the coordinates of C be (2, 2) and the coordinates of D be (1, 2). As the values of x increase, the values of y is a horizontal line. remain the same. The graph of g GH g GH slope of g GH 5 22 2 (22) 1 2 (22) 5 0 3 5 0 The slope of The slope of any horizontal line is 0. is 0. g GH No Slope. Let the coordiPoints C and D are two points on nates of C be (2, 2) and the coordinates of D be (2, 1). The values of x remain the same for all points as the values of y increase. The graph of is a vertical line. The slope of g ML is 2 2 2 5 23 22 2 1 0, which is g ML g ML undefined. g ML A vertical line has no slope. has no slope. y 1 11 G C(2, 2) O 1 x H D(1, 2) y 1 1 1 O 1 M D(2, 1) x C(2, 2) L A fundamental property of a straight line is that its slope is constant. Therefore, any two points on a line may be used to compute the slope of the line. 14365C08.pgs 7/10/07 8:46 AM Page 294 294 Slopes and Equations of Lines EXAMPLE 1 Find the slope of the line that is determined by points (2, 4) and (4, 2). Solution Plot points (2, 4) and (4, 2). Let point (2, 4) be P1(x1, y1) and let point (4, 2) be P2(x2, y2). 4, |
x2 2, y1 Then, x1 Dy Dx 5 g slope of P1P2 5 2. 4, and y2 y2 2 y1 x2 2 x1 5 2 2 4 4 2 (22) 5 22 6 5 21 3 Answer y P1(2, 4) 2 6 O 1 1 1 1 P2(4, 2) x EXAMPLE 2 Through point (1, 4), draw the line whose slope is 23 2. Solution How to Proceed (1) Graph point A(1, 4). (2) Note that, since slope Dy 3 Dx 5 2 2 5 23 2, when y decreases by 3, x increases by 2. Start at point A(1, 4) and move 3 units downward and 2 units to the right to locate point B. (3) Start at B and repeat these movements to locate point C. (4) Draw a line that passes through points A, B, and C. y 3 A(1, 4 Exercises Writing About Mathematics 1. How is the symbol y read and what is its meaning? 2. Brad said that since 0 is the symbol for “nothing,” no slope is the same as zero slope. Do you agree with Brad? Explain why or why not. 14365C08.pgs 7/10/07 8:46 AM Page 295 The Equation of a Line 295 Developing Skills In 3–11, in each case: a. Plot both points and draw the line that they determine. b. Find the slope of this line if the line has a defined slope. c. State whether the line through these points would slant upward, slant downward, be horizontal, or be vertical. 3. (0, 1) and (4, 5) 6. (1, 5) and (3, 9) 9. (1, 2) and (7, 8) 4. (1, 0) and (4, 9) 7. (5, 3) and (1, 1) 10. (4, 2) and (8, 2) 5. (0, 0) and (3, 6) 8. (2, 4) and (2, 2) 11. (1, 3) and (2, 3) In 12–23, in each case, draw a line with the given slope, m, through the given point. 12. (0, 1); m 2 15. (4, |
5); 18. (1, 3); m 1 m 5 2 3 21. (1, 0); m 5 4 Applying Skills 13. (1, 3); m 3 16. (3, 2); m 0 m 5 23 19. (2, 3); 2 22. (0, 2); m 2 3 14. (2, 5); m 1 17. (4, 7); m 2 21 20. (1, 5); m 3 23. (2, 0); m 1 2 24. a. Graph the points A(2, 4) and B(8, 4). b. From point A, draw a line that has a slope of 2. c. From point B, draw a line that has a slope of 2. d. Let the intersection of the lines drawn in b and c be C. What are the coordinates of C? of ABC. Prove that ABC is an isosce- e. Draw the altitude from vertex C to base AB les triangle. 25. Points A(3, 2) and B(9, 2) are two vertices of rectangle ABCD whose area is 24 square units. Find the coordinates of C and D. (Two answers are possible.) 26. A path to the top of a hill rises 75 feet vertically in a horizontal distance of 100 feet. Find the slope of the path up the hill. 27. The doorway of a building under construction is 3 feet above the ground. A ramp to reach the doorway is to have a slope of. How far from the base of the building should the ramp begin? 2 5 8-2 THE EQUATION OF A LINE We have learned two facts that we can use to write the equation of a line. Two points determine a line. The slope of a straight line is constant. 14365C08.pgs 7/10/07 8:46 AM Page 296 296 Slopes and Equations of Lines The second statement on the bottom of page 295 can be written as a biconditional: Postulate 8.1 A, B, and C lie on the same line if and only if the slope of slope of.BC AB is equal to the y 1 O 1 A(3, 1) B(6, 5) x Let A(3, 1) and B(6, 5) be two points g. Let P(x, y) be any other point on AB on g AB using the following fact:. We can write the equation of g |
AB by slope of AB 5 slope of BP 21 2 5 23 2 6 5 26 29 5 2 3 5 3(y 2 5) 5 2(x 2 6) 3y 2 15 5 2x 2 12 3y 5 2x 1 3 y 5 2 3x 1 1 Recall that when the equation is solved for y in terms of x, the coefficient of x is the slope of the line and the constant term is the y-intercept, the y-coordinate of the point where the line intersects the y-axis. The x-intercept is the x-coordinate of the point where the line intersects the x-axis. Procedure To find the equation of a line given two points on the line: 1. Find the slope of the line using the coordinates of the two given points. 2. Let P(x, y) be any point on the line.Write a ratio that expresses the slope of the line in terms of the coordinates of P and the coordinates of one of the given points. 3. Let the slope found in step 2 be equal to the slope found in step 1. 4. Solve the equation written in step 3 for y. 14365C08.pgs 7/10/07 8:46 AM Page 297 The Equation of a Line 297 When we are given the coordinates of one point and the slope of the line, the equation of the line can be determined. For example, if (a, b) is a point on the line whose slope is m, then the equation is: y 2 b x 2 a 5 m This equation is called the point-slope form of the equation of a line. EXAMPLE 1 The slope of a line through the point A(3, 0) is 2. a. Use the point-slope form to write an equation of the line. b. What is the y-intercept of the line? c. What is the x-intercept of the line? Solution a. Let P(x, y) be any point on the line. The slope of AP 2. y 2 0 x 2 3 5 22 y 5 22(x 2 3) y 5 22x 1 6 y P(x, y) b. The y-intercept is the y-coordinate of the point at which the line intersects the y-axis, that is, the value of y when x is 0. When x 0, 1 O 1 A(3, 0) x |
y 5 22(0) 1 6 5 0 1 6 5 6 The y-intercept is 6. When the equation is solved for y, the y-intercept is the constant term. c. The x-intercept is the x-coordinate of the point at which the line intersects the x-axis, that is, the value of x when y is 0. Since (3, 0) is a given point on the line, the x-intercept is 3. Answers a. y 2x 6 b. 6 c. 3 EXAMPLE 2 a. Show that the three points A(2, 3), B(0, 1), and C(3, 7) lie on a line. b. Write an equation of the line through A, B, and C. 14365C08.pgs 7/10/07 8:46 AM Page 298 298 Slopes and Equations of Lines Solution a. The points A(2, 3), B(0, 1), and C(3, 7) lie on the same line if and only if the slope of AB slope of AB is equal to the slope of 5 23 2 1 22 2 0 5 24 22 5 2 BC. slope of BC 5 1 2 7 0 2 3 5 26 23 5 2 The slopes are equal. Therefore, the three points lie on the same line. b. Use the point-slope form of the equation of a line. Let (x, y) be any other point on the line. You can use any of the points, A, B, or C, with (x, y) and the slope of the line, to write an equation. We will use A(2, 3). y 2 (23) x 2 (22) 5 2 y 1 3 5 2(x 1 2) y 1 3 5 2x 1 4 y 5 2x 1 1 Answers a. Since the slope of b. y 2x 1 AB is equal to the slope of BC, A, B, and C lie on a line. Alternative Solution (1) Write the slope-intercept form of an equation of a line: y mx b (2) Substitute the coordinates of A in that 3 m(2) b equation: (3) Substitute the coordinates of C in that equation: (4) Write the system of two equations from (2) and (3): 7 m(3) b 3 2m b 7 3m b ( |
5) Solve the equation 3 2m b for b in b 2m 3 terms of m: (6) Substitute the value of b found in (5) for b in the second equation and solve for m: (7) Substitute this value of m in either equation to find the value of b: The equation is y 2x 1. 7 3m b 7 3m (2m 3) 7 5m 3 10 5m 2 m b 2m 3 b 2(2) 3 b 1 14365C08.pgs 7/10/07 8:46 AM Page 299 The Equation of a Line 299 We can show that each of the given points lies on the line whose equation is y 2x 1 by showing that each pair of values makes the equation true. (2, 3) y 2x 1 2(2) 1 5? 3 3 3 ✔ (0, 1) y 2x 1 2(0) 1 5? 1 1 1 ✔ (3, 7) y 2x 1 2(3) 1 5? 7 7 7 ✔ Answers a, b: Since the coordinates of each point make the equation y 2x 1 true, the three points lie on a line whose equation is y 2x 1. Exercises Writing About Mathematics 1. Jonah said that A, B, C, and D lie on the same line if the slope of. Do you agree with Jonah? Explain why or why not. CD of AB is equal to the slope 2. Sandi said that the point-slope form cannot be used to find the equation of a line with no slope. a. Do you agree with Sandi? Justify your answer. b. Explain how you can find the equation of a line with no slope. Developing Skills In 3–14, write the equation of each line. 3. Through (1, 2) and (5, 10) 5. Through (2, 2) and (0, 6) 7. Slope 4 through (1, 1) 9. Slope 3 and y-intercept 5 11. Through (1, 5) and (4, 5) 4. Through (0, 1) and (1, 0) 6. Slope 2 and through (2, 4) 8. Slope 1 2 through (5, 4) 10. Slope 1 and x-intercept 4 12. Through (1, 5) and (1, 2) 13. x-intercept |
2 and y-intercept 4 15. a. Do the points P(3, 3), Q(5, 4), and R(1, 1) lie on the same line? 14. No slope and x-intercept 2 b. If P, Q, and R lie on the same line, find the equation of the line. If P, Q, and R do not lie on the same line, find the equations of the lines g PQ, g QR, and g.PR 14365C08.pgs 7/10/07 8:46 AM Page 300 300 Slopes and Equations of Lines 16. a. Do the points L(1, 3), M(5, 6), and N(4, 0) lie on the same line? b. If L, M, and N lie on the same line, find the equation of the line. If L, M, and N do not lie on the same line, find the equations of the lines Applying Skills g LM, g MN, and g LN. 17. At a TV repair shop, there is a uniform charge for any TV brought in for repair plus an hourly fee for the work done. For a TV that needed two hours of repair, the cost was $100. For a TV that needed one and a half hours of repair, the cost was $80. a. Write an equation that can be used to find the cost of repair, y, when x hours of work are required. Write the given information as ordered pairs, (2, 100) and (1.5, 80). b. What would be the cost of repairing a TV that requires 3 hours of work? c. What does the coefficient of x in the equation that you wrote in a represent? d. What does the constant term in the equation that you wrote in a represent? 18. An office manager buys printer cartridges from a mail order firm. The bill for each order includes the cost of the cartridges plus a shipping cost that is the same for each order. The bill for 5 cartridges was $98 and a later bill for 3 cartridges, at the same rate, was $62. a. Write an equation that can be used to find y, the amount of the bill for an order, when x cartridges are ordered. Write the given information as ordered pairs, (5, 98) and (3, 62). b. What would be the amount of the bill for 8 cartridges? c. What does the coefficient |
of x in the equation that you wrote in a represent? d. What does the constant term in the equation that you wrote in a represent? 19. Show that if the equation of the line can be written as x-axis at (a, 0) and the y-axis at (0, b). x a 1 y b 5 1, then the line intersects the 8-3 MIDPOINT OF A LINE SEGMENT The midpoint of a line segment is the point of that line segment that divides the segment into two congruent segments. In the figure, A(1, 4) and B(7, 4) determine a horizontal segment,, whose midpoint, M, can be found by first finding the distance from A to B. Since AB 7 (1) 8 units, AM 4 units, and MB 4 units. A(1, 4) B(7, 4) AB M y We can find the x-coordinate of M by adding AM to the x-coordinate of A or by subtracting MB from the x-coordinate of B. The x-coordinate of M is 1 4 3 or 7 4 3. Since A, B, and M are all on the same hori- 1 1 O 1 1 x 14365C08.pgs 7/10/07 8:46 AM Page 301 Midpoint of a Line Segment 301 zontal line, they all have the same y-coordinate, 4. The coordinates of the midpoint M are (3, 4). The x-coordinate of M is the average of the x-coordinates of A and B. x-coordinate of M 5 21 1 7 2 y CD 5 6 2 5 3 Similarly, C(3, 3) and D(3, 1) determine a vertical segment,, whose midpoint, N, can be found by first finding the distance from C to D. Since CD 1 (3) 4 units, CN 2 units, and ND 2 units. We can find the y-coordinate of N by adding 2 to the y-coordinate of C or by subtracting 2 from the y-coordinate of D. The ycoordinate of N is 3 2 1 or 1 2 1. Since C, D, and N are all on the same vertical line, they all have the same x-coordinate, 3. The coordinates of the midpoint N are (3, 1). The y-coordinate of N |
is the average of the y-coordinates of C and D. C(3, 3) D(3, 1) 1 1 O N x 1 1 y-coordinate of N 5 1 1 (23) 2 5 22 2 5 21 These examples suggest the following relationships: If the endpoints of a horizontal segment are (a, c) and (b, c), then the coordinates of the midpoint are: a 1 b 2 If the endpoints of a vertical segment are (d, e) and (d, f ), then the, c A B coordinates of the midpoint are: d, e 1 f 2 A In the figure, P(2, 1) and Q(8, 5) are the endpoints of. A horizontal line PQ through P and a vertical line through Q intersect at R(8, 1). • The coordinates of S, the midpoint of PR, are 2 1 8 2, 1 (5, 1). • The coordinates of T, the midpoint 8, 5 1 1 2 (8, 3)., are QR of A A B B B y Q(8, 5) T(8, 3) R(8, 1) x 1 O 1 P(2, 1) S(5, 1) 14365C08.pgs 7/10/07 8:46 AM Page 302 302 Slopes and Equations of Lines Now draw a vertical line through S and a horizontal line through T. These lines appear to intersect at a point on that we will call M. This point has PQ the coordinates (5, 3). We need to show that this point is a point on and is the midpoint of PQ The point M is on PQ. PQ if and only is equal to the slope if the slope of. of MQ PM y M(5, 3) 1 O 1 P(2, 1) S(5, 1) Q(8, 5) T(8, 3) R(8, 1) x slope of PM slope of MQ Since these slopes are equal, P, M, and Q lie on a line. The point M is the midpoint of if PM MQ. We can show that PQ PM MQ by showing that they are corresponding parts of congruent triangles. Therefore, • PS 5 2 3 and MT 8 5 3. PS > MT • SM 3 1 2 and TQ 5 3 2. SM > TQ Therefore,. |
. y • Since vertical lines are perpendicular to horizontal lines, PSM and MTQ are right angles and therefore congruent. 1 O 1 M(5, 3) P(2, 1) S(5, 1) Q(8, 5) T(8, 3) R(8, 1) x • Therefore, PSM MTQ by SAS and PM > MQ because correspond- ing parts of congruent triangles are congruent. We can conclude that the coordinates of the midpoint of a line segment whose endpoints are (2, 1) and (8, 5) are This example suggests the following theorem5, 3). B Theorem 8.1 If the endpoints of a line segment are (x1, y1) and nates of the midpoint of the segment are x1 1 x2 2, A y Given The endpoints of B(x2, y2). AB are A(x1, y1) and (x2, y2) y1 1 y2 2 B, then the coordi-. Prove The coordinates of the midpoint of are A x1 1 x2 2 y1 1 y2 2,. B AB O A(x1, y1) B(x2, y2) y1 y2 M(, ) 2 x1 x2 2 x 14365C08.pgs 7/10/07 8:46 AM Page 303 Midpoint of a Line Segment 303 Proof In this proof we will use the following facts from previous chapters that we have shown to be true: • Three points lie on the same line if the slope of the segment joining two of the points is equal to the slope of the segment joining one of these points to the third. • If two points lie on the same horizontal line, they have the same y-coordinate and the length of the segment joining them is the absolute value of the difference of their x-coordinates. • If two points lie on the same vertical line, they have the same x-coordinate and the length of the segment joining them is the absolute value of the difference of their y-coordinates. We will follow a strategy similar to the one used in the previous example. First, we will prove that the point M with coordinates A then we will use congruent triangles to show that tion of a midpoint of a segment, this will prove that M is the midpoint of B. From the defini- AB |
. is on AB, and, x1 1 x2 2 y1 1 y2 2 AM > MB (1) Show that M x1 1 x2 2 y1 1 y2 2, A B lies on AB : slope of AM 5 y1 1 y2 2 2 y1 2 2 x1 x1 1 x2 slope of MB 5 y2 2 x2 2 y1 1 y2 2 x1 1 x2 2 5 y1 1 y2 2 2y1 x1 1 x2 2 2x1 y2 2 y1 x2 2 x1 2y2 2 (y1 1 y2) 2x2 2 (x1 1 x2) y2 2 y1 x2 2 x1 Points A, M, and B lie on the same line because the slope of to the slope of AM MB 5 5 5. is equal (2) Let C be the point on the same vertical line as B and the same horizontal line as A. The coordinates of C are (x2, y1). y A(x1, y1) O 2 y 2 1 y 2 M(, ) 1 x 2 x B(x2, y2) y1 y2 E(x2, ) 2 x1 x2 D(, y1) 2 C(x2, y1) x The midpoint of AC is D The midpoint of BC is E. x1 1 x2, y1 B 2 y1 1 y2. 2 B A x2, A 14541C08.pgs 1/25/08 3:46 PM Page 304 304 Slopes and Equations of Lines y A(x1, y1) O 2 y 2 1 y M(, ) 1 x x 2 2 B(x2, y2) y1 y2 E(x2, ) 2 x1 x2 D(, y1) 2 C(x2, y1) x AD 5 5 P P 5 x1 1 x2 2 2 x1 P x1 1 x2 2 2x1 2 x2 2 x1 2 P P Therefore, AD = ME and P AD > ME. MD 5 5 5 P P P y1 1 y2 2 y1 2 2y1 2 y1 2 y2 2 y1 2 y2 2 P P P Therefore, MD = BE and MD > BE. ME 5 5 5 BE 5 5 5 P P P P P P x |
1 1 x2 2 x2 2 2x2 2 x1 2 x2 2 x2 2 x1 2 P y1 1 y2 2 2 y2 P y1 1 y2 2 2y2 2 y1 2 y2 2 P P P P. Therefore, ADM and MEB are right angles and are Vertical lines are perpendicular to horizontal lines. ME'BE congruent. ADM is the midpoint of MEB by SAS and. AM > MB AB >. Therefore, M x1 1 x2 2 y1 1 y2, 2 A B AD'MD and We generally refer to the formula given in this theorem, that is, x1 1 x2 2 y1 1 y2 2,, as the midpoint formula. B A EXAMPLE 1 Find the coordinates of the midpoint of the segment C(1, 5) and D(4, 1). CD whose endpoints are Solution Let (x1, y1) (1, 5) and (x2, y2) (4, 1). x1 1 x2 2 The coordinates of the midpoint are A y1 1 y2 2, B 5 5 5 A A A 21 1 4 2 3 2, 4 2 B 3 2, 2 B 5 1 (21) 2, B Answer 14365C08.pgs 7/10/07 8:46 AM Page 305 Midpoint of a Line Segment 305 EXAMPLE 2 M(1, 2) is the midpoint of coordinates of B. AB and the coordinates of A are (3, 2). Find the Solution Let the coordinates of A (x1, y1) (3, 2) and the coordinates of B (x2, y2). The coordinates of the midpoint are 23 1 x2 2 5 1 23 1 x2 5 2 x2 5 5 x1 1 x2 2, A (1, 2). y1 1 y2 2 B 2 1 y2 2 5 22 2 1 y2 5 24 y2 5 26 Answer The coordinates of B are (5, 6). EXAMPLE 3 The vertices of ABC are A(1, 1), B(7, 3), and C(2, 6). Write an equation of the line that contains the median from C to. AB Solution A median of a triangle is a line segment that joins any vertex to the midpoint of the opposite side. Let M be the midpoint of AB. (1) |
Find the coordinates of M. Let (x1, y1) be (1, 1) and (x2, y2) be (7, 3). The coordinates of M are: y C(2, 6) B(7, 3) M(4, 2) x P(x, y) 1 1 O 1 A(1, 1) 1 x1 1 x2 2 y1 1 y2 4, 2) B, 1 1 3 2 B (2) Write the equation of the line through C(2, 6) and M(4, 2). Let P(x, y) be any other point on the line. y 2 6 slope of PC 5 slope of CM 22 y 2 6 5 22(x 2 2) 2 2 4 y 2 6 5 22x 1 4 y 5 22x 1 10 Answer y 2x 10 14365C08.pgs 7/10/07 8:46 AM Page 306 306 Slopes and Equations of Lines Exercises Writing About Mathematics 1. If P(a, c) and Q(b, c) are two points in the coordinate plane, show that the coordinates of the midpoint are a 1 b 2 a 2, c A. Hint: Show that. If P(a, c) and Q(b, c) are two points in the coordinate plane, show that the coordinates of the midpoint are b 2 b 2 a 2, c A. Hint: Show that Developing Skills In 3–14, find the midpoint of the each segment with the given endpoints. 3. (1, 7), (5, 1) 6. (0, 2), (4, 6) 9. (1, 0), (0, 8) 12. (7, 2), (1, 9) In 15–20, M is the midpoint of two of the points are given. 15. A(2, 7), M(1, 6) 17. B(4, 7), M(5, 5) 19. A(3, 3), B(1, 10) Applying Skills 4. (2, 5), (8, 7) 7. (5, 1), (5, 1) 10. (3, 8), (5, 8) 13. 1 2, 3, B 1, 21 2 B A A 5. (0, 8), (10, 0) 8. (6, 6), ( |
2, 5) 11. (3, 5), (1, 1) 14. 1 3, 9, B A 2 3, 3 B A AB. Find the coordinates of the third point when the coordinates of 16. A(3, 3), M(3, 9) 18. B(4, 2), M 20. A(0, 7), M A B 3 2, 0 A 0, 7 2 B 21. The points A(1, 1) and C(9, 7) are the vertices of rectangle ABCD and B is a point on the same horizontal line as A. a. What are the coordinates of vertices B and D? b. Show that the midpoint of diagonal AC is also the midpoint of diagonal. BD 22. The points P(x1, y1) and R(x2, y2) are vertices of rectangle PQRS and Q is a point on the same horizontal line as P. a. What are the coordinates of vertices Q and S? b. Show that the midpoint of diagonal PR is also the midpoint of diagonal QS. 23. The vertices of ABC are A(1, 4), B(5, 2), and C(5, 6). a. What are the coordinates of M, the midpoint of g b. Write an equation of CM c. What are the coordinates of N, the midpoint of g BN d. Write an equation of AB? AC? that contains the median from B. that contains the median from C. 14365C08.pgs 7/10/07 8:46 AM Page 307 e. What are the coordinates of the intersection of g CM and g? BN The Slopes of Perpendicular Lines 307 f. What are the coordinates of P, the midpoint of g AP g. Write an equation of BC? that contains the median from A. g CM lie on g BN g AP and? h. Does the intersection of i. Do the medians of this triangle intersect in one point? 8-4 THE SLOPES OF PERPENDICULAR LINES Let l1 be a line whose equation is y m1x where m1 is not equal to 0. Then O(0, 0) and A(1, m1) are two points on the line. m1 2 0 slope of l1 5 1 2 0 m1 5 1 5 m1 Under a counterclockwise rotation of |
90° about the origin, the image of A(1, m1) is A(m1, 1). Since AOA is a g'OAr right angle, g OA Let l2 be the line through A(m1, 1) and O(0, 0), and let the slope of l2 be m2. Then: g OAr. y 1 l2 A(m1, 1) l1 A(1, m1 m2 5 0 2 1 5 21 0 2 (2m1) 0 1 m1 5 2 1 m1 We have shown that when two lines through the origin are perpendicular, the slope of one is the negative reciprocal of the slope of the other. Is the rule that we found for the slopes of perpendicular lines through the origin true for perpendicular lines that do not intersect at the origin? We will show this by first establishing that translations preserve slope: Theorem 8.2 Under a translation, slope is preserved, that is, if line l has slope m, then under a translation, the image of l also has slope m. 14541C08.pgs 1/25/08 3:46 PM Page 308 308 Slopes and Equations of Lines Proof: Let P(x1, y1) and Q(x2, y2) be two points on line l. Then: slope of l 5 y2 2 y1 x2 2 x1 Under a translation Ta,b, the images of P and Q have coordinates. Therefore, the slope of l, the image Pr(x1 1 a, y1 1 b) of l, is Qr(x2 1 a, y2 1 b) and slope of lr 5 5 5 (y2 1 b) 2 (y1 1 b) (x2 1 a) 2 (x1 1 a) y2 2 y1 1 b 2 b x2 2 x1 1 a 2 a y2 2 y1 x2 2 x1 As a simple application of Theorem 8.2, we can show that the slopes of any two perpendicular lines are negative reciprocals of each other: Theorem 8.3a If two non-vertical lines are perpendicular, then the slope of one is the negative reciprocal of the other. y (x1, y1) l 2 (x1 a, y1 b) O (x2 a, y2 b) (x2, y2) l 1 l2 (a, |
b) x l1 Proof: Let l1 and l2 be two perpendicular lines that intersect at (a, b). Under the translation (x, y) → (x a, y b), the image of (a, b) is (0, 0). l2r l1r l1r and Theorem 8.2 tells us that if the slope of l1 is m, then the slope of its, is m. Since l1 and l2 are perimage, pendicular, their images,, are also perpendicular because translations preserve angle measure. Using what we established at the beginning of the section, since the slope of is l2r m, the slope of. Slope is preis served under a translation. Therefore, 21. the slope of l2 is m 21 m l1r The proof of Theorem 8.3a is called a transformational proof since it uses transformations to prove the desired result. Is the converse of Theorem 8.3a also true? To demonstrate that it is, we need to show that if the slope of one line is the negative reciprocal of the slope of the other, then the lines are perpendicular. We will use an indirect proof. 14365C08.pgs 7/10/07 8:46 AM Page 309 Theorem 8.3b If the slopes of two lines are negative reciprocals, then the two lines are perpendicular. The Slopes of Perpendicular Lines 309 Given Two lines, g AB, that intersect at A. The slope of is m and the slope g AC of is 21 m, the negative reciprocal g AC and g AB of m. Prove g AB g'AC C D B A y O x Proof Statements Reasons g AC 1. is not perpendicular g. AB to 1. Assumption. g AD perpendicular 2. At a point on a given line, one and 2. Construct g AB to at A. 3. Slope of g AB 4. The slope of is m. g AD is 21 m. only one perpendicular can be drawn. 3. Given. 4. If two lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other. 5. The slope of g AC is 21 m. 5. Given. 6. A, C, and D are on the same g AC g AD and are line, that is, the same line. 6. Three points lie on the same line if and only if the slope of the segment joining two |
of the points is equal to the slope of a segment joining another pair of these points. g AC g'AB 7. 7. Contradiction (steps 1, 6). 14365C08.pgs 7/10/07 8:46 AM Page 310 310 Slopes and Equations of Lines We can restate Theorems 8.3a and 8.3b as a biconditional. Theorem 8.3 Two non-vertical lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other. EXAMPLE 1 What is the slope of l2, the line perpendicular to l1, if the equation of l1 is x 2y 4? Solution (1) Solve the equation of l1 for y: x 1 2y 5 4 2y 5 2x 1 4 y 5 21 y 5 21 2x 1 2 2x 1 2 slope slope of l2 5 2 (2) Find the slope of l1: (3) Find the slope of l2, the negative reciprocal of the slope of l1: Answer The slope of l2 is 2. EXAMPLE 2 Show that ABC is a right triangle if its vertices are A(1, 1), B(4, 3), and C(2, 6). Solution The slope of AB is 23 2 5 23 2 1 2 6 1 2 2 5 25 21 5 5.. The slope of BC is The slope of AC The slope of BC is is the negative reciprocal of the slope of. Therefore,, B is a right angle, and ABC is a right triangle. g AB AB g BC is perpendicular to Exercises Writing About Mathematics 1. Explain why the slope of a line perpendicular to the line whose equation is x 5 cannot be found by using the negative reciprocal of the slope of that line. 2. The slope of a line l is 21 3 1. Kim said that the slope of a line perpendicular to l is. 1 3 Santos said that the slope of a line perpendicular to l is 3. Who is correct? Explain your answer. 14365C08.pgs 7/10/07 8:46 AM Page 311 The Slopes of Perpendicular Lines 311 Developing Skills In 3–12: a. Find the slope of the given line. b. Find the slope of the line perpendicular to the given line. 3. y 4x 7 5. x y 8 7. 3x 5 2 |
y 9. through (0, 4) and (2, 0) 11. through (4, 4) and (4, 2) 4. y x 2 6. 2x y 3 8. through (1, 1) and (5, 3) 10. y-intercept 2 and x-intercept 4 12. parallel to the x-axis through (5, 1) In 13–16, find the equation of the line through the given point and perpendicular to the given line. 13. 21 2, 22 ; 2x 7y 15 14. (0, 0); 2x 4y 12 B y 5 21 A 15. (7, 3); 17. Is the line whose equation is y 3x 5 perpendicular to the line whose equation is 16. (2, 2); y 1 3x 2 3 3x y 6? 18. Two perpendicular lines have the same y-intercept. If the equation of one of these lines is y x 1, what is the equation of the other line? 1 2 19. Two perpendicular lines intersect at (2, 1). If x y 3 is the equation of one of these lines, what is the equation of the other line? 20. Write an equation of the line that intersects the y-axis at (0, 1) and is perpendicular to the line whose equation is x + 2y 6. In 21–24, the coordinates of the endpoints of a line segment are given. For each segment, find the equation of the line that is the perpendicular bisector of the segment. 21. A(2, 2), B(21, 1) 23. A(3, 9), B(3, 9) Applying Skills 22. A 21 2, 3, B 3 2, 1 A B 24. A(4, 1), B(3, 3) A B 25. The vertices of DEF triangle are D(3, 4), E(1, 2), and F(3, 2). a. Find an equation of the altitude from vertex D of DEF. b. Is the altitude from D also the median from D? Explain your answer. c. Prove that DEF is isosceles. 26. If a four-sided polygon has four right angles, then it is a rectangle. Prove that if the vertices of a polygon are A(3, 2), B(5, 1), C(1, 5), and D(3, 2 |
), then ABCD is a rectangle. 14365C08.pgs 7/10/07 8:46 AM Page 312 312 Slopes and Equations of Lines 27. The vertices of ABC are A(2, 2), B(6, 6) and C(6, 0). a. What is the slope of AB? b. Write an equation for the perpendicular bisector of AB. c. What is the slope of BC? d. Write an equation for the perpendicular bisector of BC. e. What is the slope of AC? f. Write an equation for the perpendicular bisector of g. Verify that the perpendicular bisectors of ABC intersect in one point. 28. The coordinates of the vertices of ABC are A(2, 0), B(4, 0), and C(0, 4). AC. a. Write the equation of each altitude of the triangle. b. Find the coordinates of the point of intersection of these altitudes. 29. The coordinates of LMN are L(2, 5), M(2, 3), and N(7, 3). a. Using the theorems of Section 7-6, prove that L and N are acute angles. b. List the sides of the triangles in order, starting with the shortest. c. List the angles of the triangles in order, starting with the smallest. Hands-On Activity The following activity may be completed using graph paper, pencil, compass, and straightedge, or geometry software. In the exercises of Section 6-5, we saw how a translation in the horizontal direction can be achieved by a composition of two line reflections in vertical lines and a translation in the vertical direction can be achieved by a composition of two line reflections in horizontal lines. In this activity, we will see how any translation is a composition of two line reflections.. STEP 1. Draw any point A on a coordinate plane. STEP 2. Translate the point A to its image A under the given translation, Ta,b. g AAr STEP 3. Draw line STEP 4. Draw any line l1 perpendicular to STEP 5. Reflect the point A in line l1. Let its image be A. STEP 6. Let l2 be the line that is the perpendicular bisector of Result: The given translation, Ta,b + rl2 Ta,b 5 rl1 order, that is,, is the composition of the two line reflections,. (Recall that is performed first.) g AAr Ar |
As rl1.. rl1 and rl2, in that For a–c, using the procedure above write the equations of two lines under which reflections in the two lines are equal to the given translation. Check your answers using the given coordinates. a. T4,4 c. T1,3 b. T3,2 D(0, 0), E(5, 3), F(2, 2) D(1, 2), E(5, 3), F(5, 6) D(6, 6), E(2, 5), F(1, 6) 14365C08.pgs 7/10/07 8:46 AM Page 313 8-5 COORDINATE PROOF Coordinate Proof 313 Many of the proofs that we did in the preceding chapters can also be done using coordinates. In particular, we can use the formula for slope, the slopes of perpendicular lines, and the coordinates of the midpoint of a line segment presented in this chapter to prove theorems about triangles. In later chapters we will use coordinates to prove theorems about polygons, parallel lines, and distances. There are two types of proofs in coordinate geometry: 1. Proofs Involving Special Cases. When the coordinates of the endpoints of a segment or the vertices of a polygon are given as ordered pairs of numbers, we are proving something about a specific segment or polygon. (See Example 1.) 2. Proofs of General Theorems. When the given information is a figure that represents a particular type of polygon, we must state the coordinates of its vertices in general terms using variables. Those coordinates can be any convenient variables. Since it is possible to use a transformation that is an isometry to move a triangle without changing its size and shape, a geometric figure can be placed so that one of its sides is a segment of the x-axis. If two line segments or adjacent sides of a polygon are perpendicular, they can be represented as segments of the x-axis and the y-axis. (See Example 2.) To prove that line segments bisect each other, show that the coordinates of the midpoints of each segment are the same ordered pair, that is, are the same point. To prove that two lines are perpendicular to each other, show that the slope of one line is the negative reciprocal of the slope of the other. The vertices shown in the diagrams below can be used when working with triangles. y ( |
0, b) y (0, b) y (0, b) y (0, b) (a, 0) O (c, 0) x O (a, 0) (c, 0) x (0, 0) (a, 0) x (a, 0) O (a, 0) x The triangle with vertices (a, 0), (0, b), (c, 0) can be any triangle. It is convenient to place one side of the triangle on the x-axis and the vertex opposite that side on the y-axis. The triangle can be acute if a and c have opposite signs or obtuse if a and c have the same sign. A triangle with vertices at (a, 0), (0, 0), (0, b) is a right triangle because it has a right angle at the origin. 14365C08.pgs 7/10/07 8:46 AM Page 314 314 Slopes and Equations of Lines A triangle with vertices at (a, 0), (0, b), (a, 0) is isosceles because the alti- tude and the median are the same line segment. When a general proof involves the midpoint of a segment, it is helpful to express the coordinates of the endpoints of the segment as variables divisible by 2. For example, if we had written the coordinates of the vertices of a right triangle as (d, 0), (0, e) and ( f, 0), we could simply let d 2a, e 2b, and f 2c so that the coordinates would be (2a, 0), (0, 2b) and (2c, 0). The coordinates of midpoints of the sides of this triangle would be simpler using these coordinates. EXAMPLE 1 AB and bisect each other and are perpendicular to each other Prove that if the coordinates of the endpoints of these segments are A(3, 5), B(5, 1), C(2, 3), and D(4, 9). CD Solution This is a proof involving a special case. B 5 1 1 2 23 1 5 2 The midpoint of, A The midpoint of, 23 1 9 2 22 1 4 2 A The slope of B AB The slope of CD is AB 2 5 2, 6 2 B 5 (1, 3). 5 (1, 3). A is CD 6 2 5 2, 2 |
B A 5 2 1 23 2 5 5 4 9 2 (23) 4 2 (22) 5 12 is is 28 5 21 2. 6 5 2. y A(3, 5) D(4, 9) (1, 3) 1 O 1 C(2, 3) B(5, 1) x CD CD and and AB AB rocal of the slope of the other. bisect each other because they have a common midpoint, (1, 3). are perpendicular because the slope of one is the negative recip- EXAMPLE 2 Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices. Given: Right triangle ABC whose vertices are A(2a, 0), B(0, 2b), and C(0, 0). Let M be the midpoint of the hypotenuse. AB Prove: AM BM CM y B(0, 2b) M(a, b) C(0, 0) N(a, 0) A(2a, 0) x 14365C08.pgs 7/10/07 8:46 AM Page 315 Coordinate Proof 315 Proof This is a proof of a general theorem. Since it is a right triangle, we can place one vertex at the origin, one side of the triangle on the x-axis, and a second side on the y-axis so that these two sides form the right angle. We will use coordinates that are divisible by 2 to simplify computation of midpoints. (1) The midpoint of a line segment is a point of that line segment that sepa- rates the line segment into two congruent segments. Therefore, AM > BM. (2) The coordinates of M are 2a 1 0 2, 0 1 2b 2 (a, b). B A (3) From M, draw a vertical segment that intersects AC at N. The x-coordi- nate of N is a because it is on the same vertical line as M. The y-coordinate of N is 0 because it is on the same horizontal line as A and C. The coordinates of N are (a, 0). (4) The midpoint of AC is and AN > NC. 0 1 2a a, 0). N is the midpoint of AC (5) The vertical segment MN is perpendicular to the horizontal segment AC. Perpendicular lines are two lines that intersect to form right angles |
. Therefore, ANM and CMN are right angles. All right angles are congruent, so ANM CMN. Also, MN > MN. (6) Then, AMN CMN by SAS (steps 4 and 5). (7) AM > CM because they are corresponding parts of congruent triangles. (8) (step 1) and AM > BM AM > BM > CM or AM BM = CM. The midpoint of the hypotenuse of a right triangle is equidistant from the vertices. (step 7). Therefore, AM > CM Exercises Writing About Mathematics 1. Ryan said that (a, 0), (0, b), (c, 0) can be the vertices of any triangle but if b 5 2b a c, then the triangle is a right triangle. Do you agree with Ryan? Explain why or why not. 2. Ken said that (a, 0), (0, b), (c, 0) can be the vertices of any triangle but if a and c have the same sign, then the triangle is obtuse. Do you agree with Ken? Explain why or why not. Developing Skills 3. The coordinates of the endpoints of AB are A(0, 2) and B(4, 6). The coordinates of the endpoints of segments bisect each other. CD are C(4, 5) and D(8, 1). Using the midpoint formula, show that the line 14365C08.pgs 7/10/07 8:46 AM Page 316 316 Slopes and Equations of Lines 4. The vertices of polygon ABCD are A(2, 2), B(5, 2), C(9, 1), and D(6, 5). Prove that the diagonals AC and BD are perpendicular and bisect each other using the midpoint formula. 5. The vertices of a triangle are L(0, 1), M(2, 5), and N(6, 3). a. Find the coordinates K, the midpoint of the base, LN. MK is an altitude from M to b. Show that c. Using parts a and b, prove that LMN is isosceles. 6. The vertices of ABC are A(1, 7), B (9, 3), and C(3, 1). LN. a. Prove that ABC is a right triangle. b. Which angle is the |
right angle? c. Which side is the hypotenuse? d. What are the coordinates of the midpoint of the hypotenuse? e. What is the equation of the median from the vertex of the right angle to the hypotenuse? f. What is the equation of the altitude from the vertex of the right angle to the hypotenuse? g. Is the triangle an isosceles right triangle? Justify your answer using parts e and f. 7. The coordinates of the vertices of ABC are A(4, 0), B(0, 8), and C(12, 0). a. Draw the triangle on graph paper. b. Find the coordinates of the midpoints of each side of the triangle. c. Find the slope of each side of the triangle. d. Write the equation of the perpendicular bisector of each side of the triangle. e. Find the coordinates of the circumcenter of the triangle. 8. A rhombus is a quadrilateral with four congruent sides. The vertices of rhombus ABCD are A(2, 3), B(5, 1), C(10, 1), and D(7, 3). a. Prove that the diagonals AC and BD bisect each other. b. Prove that the diagonals AC and BD are perpendicular to each other. Applying Skills 9. The vertices of rectangle PQRS are P(0, 0), Q(a, 0), R(a, b), and S(0, b). Use congruent tri- angles to prove that the diagonals of a rectangle are congruent, that is, PR > QS. 10. The vertices of square EFGH are E(0, 0), F(a, 0), G(a, a), and H(0, a). Prove that the diago- nals of a square, point formula. EG and FH, are the perpendicular bisectors of each other using the mid- 11. Use congruent triangles to prove that (0, 0), (2a, 0), and (a, b) are the vertices of an isosce- les triangle. (Suggestion: Draw the altitude from (a, b).) 14365C08.pgs 7/10/07 8:46 AM Page 317 Concurrence of the Altitudes of a Triangle 317 12. Use a translation to prove that (a |
, 0), (0, b), and (a, 0) are the vertices of an isosceles tri- angle. (Hint: A translation will let you use the results of Exercise 11.) 13. The coordinates of the vertices of ABC are A(0, 0), B(2a, 2b), and C(2c, 2d). a. Find the coordinates of E, the midpoint of AB and of F, the midpoint of AC. b. Prove that the slope of EF 14. The endpoints of segment AB is equal to the slope of are (a, 0) and (a, 0).. BC a. Use congruent triangles to show that P(0, b) and Q(0, c) are both equidistant from the endpoints of g PQ b. Show that. AB is the perpendicular bisector of. AB 8-6 CONCURRENCE OF THE ALTITUDES OF A TRIANGLE The postulates of the coordinate plane and the statements that we have proved about the slopes of perpendicular lines make it possible for us to prove that the three altitudes of a triangle are concurrent, that is, they intersect in one point. Theorem 8.4 The altitudes of a triangle are concurrent. Proof: We can place the triangle anywhere in the coordinate plane. We will place lies on the x-axis and B lies on the y-axis. Let A(a, 0), B(0, b), and it so that C(c, 0) be the vertices of ABC. Let be the altitude from B to be the altitude from A to be the altitude from C to, and. AB, BC AE AC AC BO CF y B(0, b) E F y B(0, b) F E y F E B(0, b) A(a, 0) O C(c, 0) x A(a, 0) O C(c, 0) x x A(a, 0) O C(c, 0) x B is acute B is right B is obtuse In the figures, A and C are acute angles. 14365C08.pgs 7/10/07 8:46 AM Page 318 318 Slopes and Equations of Lines y B(0, b) E F A(a, 0) O C(c, 0) x y B(0, b) F E A(a, |
0) O C(c, 0) x y F E We will show that altitudes. and BO CF tudes AE and BO intersect in the same point as alti- Intersection of altitudes and AE BO Intersection of altitudes and BO CF 1. The slope of side 0 2 b c 2 0 5 2b c. BC is 2. The slope of altitude is perpendicular to BC AE, which c, is. b 1. The slope of side 0 2 b a 2 0 5 2b a. AB is 2. The slope of altitude is perpendicular to AB CF, which a, is. b 3. The equation of AE is 3. The equation of CF is ac or ac a or y x. b b AC is a vertical line, a segment is a horizontal line, 4. Since BO of the y-axis since B is on the y-axis. AC is a vertical line, a segment is a horizontal line, 4. Since BO of the y-axis since B is on the y-axis. B(0, b) 5. The equation of g BO is x 0. 5. The equation of g BO is x 0. A(a, 0) O C(c, 0) x AE 6. The coordinates of the and intersection of be found by finding the common solution of their c equations: y x and x 0. b can BO ac b CF 6. The coordinates of the and intersection of be found by finding the common solution of their a equations: y x and x 0. b can BO ac b 7. Since one of the equations is x 0, replace x by 0 in the other equation: 7. Since one of the equations is x 0, replace x by 0 in the other equation: bx 2 ac b b(0) 2 ac b y 5 c 5 c 5 2ac b 8. The coordinates of the AE and BO intersection of 0, 2ac b B are A. bx 2 ac b b(0) 2 ac b y 5 a 5 a 5 2ac b 8. The coordinates of the and CF intersection of 0, 2ac are b B A. BO The altitudes of a triangle are concurrent at 0, 2ac b B is the altitude from C to A. Note: If B is a right angle, from A to CB three altitudes is B(0, b)., and BO CB is the altitude from B to AC AB, AB is the altitude. The intersection of these |
14365C08.pgs 7/10/07 8:46 AM Page 319 Concurrence of the Altitudes of a Triangle 319 The point where the altitudes of a triangle intersect is called the orthocenter. EXAMPLE 1 The coordinates of the vertices of PQR are P(0, 0), Q(2, 6), and R(4, 0). Find the coordinates of the orthocenter of the triangle. Solution Let PL be the altitude from P to. QR The slope of QR is 6 2 0 22 2 4 5 6 26 5 21. The slope of PL is 1. The equation of y x. PL is y 2 0 x 2 0 5 1 or QN be the altitude from Q to Let PR The point of intersection, N, is on the line determined by P and R.. PR The slope of zontal line. Therefore, of a vertical line that has no slope. is 0 since QN is a horiPR is a segment The equation of QN is x 2. y Q(2, 6) L N S(2, 2) P(0, 0) M R(4, 0) is the common solution of the The intersection S of the altitudes equations x 2 and y x. Therefore, the coordinates of the intersection S are (2, 2). By Theorem 8.4, point S is the orthocenter of the triangle or the point where the altitudes are concurrent. and QN PL Answer The orthocenter of PQR is S(2, 2). Alternative Solution Use the result of the proof given in this section. The coordinates of the point of intersection of 0, 2ac the altitudes are b B to apply, Q must lie on the y-axis and P and R must lie on the x-axis. Since P and R already lie on the x-axis, we need only to use a translation to move Q in the horizontal direction to the y-axis. Use the translation (x, y) → (x + 2, y):. In order for this result A y Q(2, 6) Q(0, 6) R(6, 0) x P(2, 0) R(4, 0) P(0, 0) P(0, 0) → P(2, 0) Q(2, 6) → Q(0, 6) R(4, 0) → R(6, 0) Therefore, A |
(a, 0) P(2, 0) or a 2 B(0, b) Q(0, 6) or b 6 C(c, 0) R(6, 0) or c 6. 14365C08.pgs 7/10/07 8:46 AM Page 320 320 Slopes and Equations of Lines The coordinates of S, the point at which the altitudes of PRQ intersect, are 0, 22(6) 6 B The intersection of the altitudes of PQR is S, the preimage of S(0, 2) under the translation (x, y) → (x + 2, y). Therefore, the coordinates of S are (2, 2). 0, 2ac b B (0, 2) 5 A A Answer The orthocenter of PQR is (2, 2). EXAMPLE 2 The coordinates of the vertices of ABC are A(0, 0), B(3, 4) and C(2, 1). Find the coordinates of the orthocenter of the triangle. Solution The slope of AC is 1 2 0 2 2 0 5 1 2. Let BD be the altitude from B to. AC The slope of altitude is 2. BD g BD The equation of line 4 2 y 3 2 x 5 22 4 2 y 5 22(3 2 x) is 2y 5 26 1 2x 2 4 y 5 22x 1 10 4 2 1 3 2 2 5 3 is BC 1 5 3. be the altitude from A to. BC The slope of Let AE The slope of altitude The equation of line AE g AE is 21 3. is B(3, 4) D C(2, 1) A(0, 0) E S(6, 2) y 2 0 x 2 0 5 21 3 y 5 21 3x The orthocenter S is the common solution of the equations y 2x 10 and y 5 21. 3x 22x 1 10 5 21 3x 212 3x 5 210 25 3x 5 210 x 5 210 x 5 6 21 23 5 B A The x-coordinate is 6; the y-coordinate is 3(6) 5 22. Answer The coordinates of the orthocenter are (6, 2) 14365C08.pgs 7/10/07 8:46 AM Page 321 Concurrence of the Altitudes of a Triangle 321 Exercises Writing About Mathematics In 1–2, the vertices of ABC are A |
(a, 0), B(0, b), and C(c, 0), as shown in the diagrams of the proof of Theorem 8.4. Assume that b 0. 1. Esther said that if A is to the left of the origin and C is to the right of the origin, then the point of intersection of the altitudes is above the origin. Do you agree with Esther? Explain why or why not. 2. Simon said that if A is an obtuse angle, and both A and C are to the right of the origin, then the point of intersection of the altitudes is above the origin. Do you agree with Simon? Explain why or why not. Developing Skills 3. The coordinates of DEF are D(9, 0), E(0, 12), and F(16, 0). a. Show that DEF is a right triangle. b. Show that E is the orthocenter of the triangle. In 4–9, find the coordinates of the orthocenter of each triangle with the given vertices. 4. A(2, 0), B(0, 6), C(3, 0) 6. L(7, 2), M(2, 12), N(11, 2) 8. G(5, 2), H(4, 8), I(5, 1) 5. D(12, 0), E(0, 8), F(6, 0) 7. P(3, 4), Q(1, 8), R(3, 4) 9. J(0, 3), K(3, 4), L(2, 1) Applying Skills 10. Two of the vertices of ABC are A(3, 0) and C(6, 0). The altitudes from these vertices intersect at P(0, 3). g AB g CB a. b. is a line through A, perpendicular to CP. Write the equation of is a line through C, perpendicular to AP. Write the equation of g. AB g CB. c. Find B, the intersection of g AB and g CB. d. Write an equation of the line that contains the altitude from B to. AC e. Show that P is a point on that line. 11. Two of the vertices of ABC are A(2, 2) and C(5, 5). The altitudes from these vertices intersect at P(1, 1). g AB g CB a. b. is |
a line through A, perpendicular to CP. Write the equation of is a line through C, perpendicular to AP. Write the equation of g AB g CB.. c. Find B, the intersection of g AB and g.CB 14365C08.pgs 8/2/07 5:48 PM Page 322 322 Slopes and Equations of Lines d. Write an equation of the line that contains the altitude from B to. AC e. Show that P is a point on that line. Hands-On Activity In this activity we will use a compass and a straightedge to construct the orthocenters of various triangles. For each triangle: a. Graph the triangle on paper or using geometry software. b. Using a compass, straightedge, and pencil, or geometry software, construct the orthocenter. (If using the computer, you are only allowed to use the point, line segment, line, and circle creation tools of your geometry software and no other software tools.) (1) A(3, 0), B(0, 2), C(4, 0) (2) D(4, 7), E(0, 5), F(3, 1) (3) G(4, 2), H(6, 0), I(0, 4) CHAPTER SUMMARY Definitions to Know • The slope, m, of a line that passes through any two points A( ) and, is the ratio of the difference of the y-values of ), where B( these points to the difference of the corresponding x-values. 2 x2 x1, y1 x2, y2 x1 slope of g AB m y2 2 y1 x2 2 x1 • Three or more lines are concurrent if they intersect in one point. • The orthocenter of a triangle is the point at which the altitudes of a trian- gle intersect. Postulate 8.1 A, B, and C lie on the same line if and only if the slope of AB is equal to the slope of BC. Theorems 8.1 If the endpoints of a line segment are, then the coordi- (x1,y1) and x1 1 x2 2, (x2,y2) y1 1 y2 2 A. B nates of the midpoint of the segment are 8.2 Under a translation, slope is preserved, that is, if line l has slope m, then under a translation, the image of l |
also has slope m. 8.3 Two non-vertical lines are perpendicular if and only if the slope of one is the negative reciprocal of the other. 8.4 The altitudes of a triangle are concurrent. 14365C08.pgs 7/10/07 8:46 AM Page 323 Chapter Summary 323 VOCABULARY 8-1 Slope • x • y 8-2 y-intercept • x-intercept • Point-slope form of an equation • Slope- intercept form of an equation 8-3 Midpoint formula 8-4 Transformational proof 8-6 Orthocenter REVIEW EXERCISES In 1–3, the coordinates of the endpoints of a line segment are given. a. Find the coordinates of each midpoint. b. Find the slope of each segment. c. Write an equation of the line that is determined by each pair of points. 1. (0, 0) and (6, 4) 4. The coordinates of P are (2, 5) and the coordinates of Q are (6, 1). 2. (3, 2) and (7, 4) 3. (1, 1) and (5, 5) a. What is the slope of g PQ b. What is the equation of? g PQ? c. What are the coordinates of the midpoint of? PQ d. What is the equation of the perpendicular bisector of 5. The vertices of RST are R(2, 2), S(1, 4), and T(7, 1).? PQ a. Show that RST is a right triangle. b. Find the coordinates of the midpoint of. RT c. Write the equation of the line that contains the median from S. d. Show that the median of the triangle from S is also the altitude from S. e. Prove that RST is an isosceles triangle. 6. Two of the vertices of ABC are A(1, 2) and B(9, 6). The slope of AC is 1 and the slope of 1 3 7. The vertices of DEF are D(1, 1), E(5, 5), F(1, 5). is. What are the coordinates of B? BC a. Find the coordinates of the midpoint of each side of the triangle. b. Find the slope of each side of the triangle. c. Find the slope of each altitude of the triangle. d |
. Write an equation of the perpendicular bisector of each side of the tri- angle. e. Show that the three perpendicular bisectors intersect in a point and find the coordinates of that point. 14365C08.pgs 7/10/07 8:46 AM Page 324 324 Slopes and Equations of Lines 8. The vertices of ABC are A(7, 1), B(5, 3), and C(3, 5). a. Prove that ABC is a right triangle. b. Let M be the midpoint of AB AMN CMN and use this result to show that M is equidistant from the vertices of ABC. and N be the midpoint of AC. Prove that 9. The vertices of DEF are D(2, 3), E(5, 0), and F(2, 3) a. Find the coordinates of M, the midpoint of DF. b. Show that DE > FE. 10. The coordinates of the vertices of quadrilateral ABCD are A(5, 4), B(1, 6), C(1, 2), and D(4, 1). Show that ABCD has a right angle. Exploration The following exploration may be completed using graph paper or using geometry software. We know that the area of a triangle is equal to one-half the product of the lengths of the base and height. We can find the area of a right triangle in the coordinate plane if the base and height are segments of the x-axis and y-axis. The steps that follow will enable us to find the area of any triangle in the coordinate plane. For example, find the area of a triangle if the vertices have the coordinates (2, 5), (5, 9), and (8, 2). STEP 1. Plot the points and draw the triangle. STEP 2. Through the vertex with the smallest x-coordinate, draw a vertical line. STEP 3. Through the vertex with the largest x-coordinate, draw a vertical line. STEP 4. Through the vertex with the smallest y-coordinate, draw a horizontal line. STEP 5. Through the vertex with the largest y-coordinate, draw a horizontal line. STEP 6. The triangle is now enclosed by a rectangle with horizontal and vertical sides. Find the length and width of the rectangle and the area of the rectangle. STEP 7. The rectangle is separated into four triangles: the given triangle and three triangles that have a base and altitude that |
are a horizontal and a vertical line. Find the area of these three triangles. STEP 8. Find the sum of the three areas found in step 7. Subtract this sum from the area of the rectangle. The difference is the area of the given triangle. Repeat steps 1 through 8 for each of the triangles with the given vertices. a. (2, 0), (3, 7), (6, 1) b. (3, 6), (0, 4), (5, 1) c. (0, 5), (6, 6), (2, 5) Can you use this procedure to find the area of the quadrilateral with vertices at (0, 2), (5, 2), (5, 3), and (2, 5)? 14541C08.pgs 1/25/08 3:46 PM Page 325 CUMULATIVE REVIEW Part I Cumulative Review 325 Chapters 1–8 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. When the coordinates of A are (2, 3) and of B are (2, 7), AB equals (1) 4 (2) 4 (3) 10 (4) 10 2. The slope of the line whose equation is 2x y 4 is (1) 1 2 (2) 2 (3) 2 (4) 4 3. Which of the following is an example of the associative property for multi- plication? (1) 3(2 5) 3(5 2) (2) 3(2 5) (3 2) 5 4. The endpoints of midpoint of (1) (2, 3) AB AB are (3) 3(2 5) 3(2) 5 (4) 3 (2 5) (3 2) 5 are A(0, 6) and B(4, 0). The coordinates of the (2) (2, 3) (3) (2, 3) (4) (2, 3) 5. In isosceles triangle DEF, DE = DF. Which of the following is true? (1) D E (2) F E (3) D F (4) D E F 6. The slope of line l is. The slope of a line perpendicular to l is (1) 2 3 22 3 (3) 3 2 (4) 23 2 2 3 (2) 7. The coordinates of two points are |
(0, 6) and (3, 0). The equation of the line through these points is (1) y 2x 6 (2) y 2x 6 1 (3) y 3 2x 1 (4) y x 3 2 8. The converse of the statement “If two angles are right angles then they are congruent” is (1) If two angles are congruent then they are right angles. (2) If two angles are not right angles then they are not congruent. (3) Two angles are congruent if and only if they are right angles. (4) Two angles are congruent if they are right angles. 9. The measure of an angle is twice the measure of its supplement. The mea- sure of the smaller angle is (1) 30 (2) 60 (3) 90 (4) 120 10. Under a reflection in the y-axis, the image of (4, 2) is (2) (4, 2) (3) (4, 2) (1) (4, 2) (4) (2, 4) 14365C08.pgs 7/10/07 8:46 AM Page 326 326 Slopes and Equations of Lines 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. a. Draw ABC in the coordinate plane if the coordinates of A are (3, 1), of B are (1, 5), and of C are (5, 2). b. Under a translation, the image of A is A(1, 1). Draw ABC, the image of ABC under this translation, and give the coordinates of B and C. c. If this translation can be written as (x, y) → (x a, y b), what are the values of a and b? 12. The coordinates of the vertices of DEF are D(1, 6), E(3, 3), and F(1, 2). Is DEF a right triangle? Justify your answer. 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 the diagram, TQ'RS h TQ bisects RTS and. Prove that RST is isosceles. 14. The following statements are true: • If Evanston is not the capital of Illinois, then Chicago is not the capital. • Springfield is the capital of Illinois or Chicago is the capital of Illinois. T • Evanston is not the capital of Illinois. Use the laws of logic to prove that Springfield is the capital of Illinois. R Q S 14365C08.pgs 7/10/07 8:46 AM Page 327 Cumulative Review 327 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. Find the coordinates of the point of intersection of the lines whose equa- tions are y 3x 1 and x 2y 5. 16. Given: CGF and DGE bisect each other at G. Prove: CD > FE 14365C09.pgs 7/10/07 8:48 AM Page 328 CHAPTER 9 CHAPTER TABLE OF CONTENTS 9-1 Proving Lines Parallel 9-2 Properties of Parallel Lines 9-3 Parallel Lines in the Coordinate Plane 9-4 The Sum of the Measures of the Angles of a Triangle 9-5 Proving Triangles Congruent by Angle, Angle, Side 9-6 The Converse of the Isosceles Triangle Theorem 9-7 Proving Right Triangles Congruent by Hypotenuse, Leg 9-8 Interior and Exterior Angles of Polygons Chapter Summary Vocabulary Review Exercises Cumulative Review 328 PARALLEL LINES “If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.” This statement, Euclid’s fifth postulate, is called Euclid’s parallel postulate. Throughout history this postulate has been questioned by mathematicians because many felt it was too complex to be a postulate. Throughout the history of mathematics, attempts were made to prove this postulate or state a related |
postulate that would make it possible to prove Euclid’s parallel postulate. Other postulates have been proposed that appear to be simpler and which could provide the basis for a proof of the parallel postulate. The form of the parallel postulate most commonly used in the study of elementary geometry today was proposed by John Playfair (1748–1819). Playfair’s postulate states: Through a point not on a given line there can be drawn one and only one line parallel to the given line. 14365C09.pgs 7/10/07 8:48 AM Page 329 9-1 PROVING LINES PARALLEL Proving Lines Parallel 329 You have already studied many situations involving intersecting lines that lie in the same plane. When all the points or lines in a set lie in a plane, we say that these points or these lines are coplanar. Let us now consider situations involving coplanar lines that do not intersect in one point. DEFINITION Parallel lines are coplanar lines that have no points in common, or have all points in common and, therefore, coincide. The word “lines” in the definition means straight lines of unlimited extent. We say that segments and rays are parallel if the lines that contain them are parallel. We indicate that g CD. The parallel lines is parallel to g CD g AB and g AB g AB g CD by writing extended indefi- nitely never intersect and have no points in common. The parallel lines g AB and g CD may have all points in common, that is, be two different names for the same line. A line is parallel to itself. Thus, g AB g CD. In Chapter 4, we stated the following postulate: g AB g AB, g CD g CD and C A D B A B C D Two distinct lines cannot intersect in more than one point. This postulate, together with the definition of parallel lines, requires that g AB and g : CD one of three possibilities exist for any two coplanar lines, 1. 2. 3. g AB g AB g AB g AB g AB g AB and and and and and and g CD g CD g CD g CD g CD g CD have no points in common. are parallel. have only one point in common. intersect. have all points in common. are the same line. These three possibilities can also be stated in the following postulate: Postulate 9.1 Two distinct coplanar lines are either parallel |
or intersecting. 14365C09.pgs 7/10/07 8:48 AM Page 330 330 Parallel Lines EXAMPLE 1 If line l is not parallel to line p, what statements can you make about these two lines? Solution Since l is not parallel to p, l and p cannot be the same line, and they have exactly one point in common. Answer Parallel Lines and Transversals g g AB intersects CD C A B D When two lines intersect, four angles are formed that have the same vertex and no common interior points. In this set of four angles, there are two pair of congruent vertical angles and four pair of supplementary adjacent angles. When two lines are intersected by a third line, two such sets of four angles are formed. DEFINITION A transversal is a line that intersects two other coplanar lines in two different points Two lines, l and m, are cut by a transversal, t. Two sets of angles are formed, each containing four angles. Each of these angles has one ray that is a subset of l or of m and one ray that is a subset of t. In earlier courses, we learned names to identify these sets of angles. • The angles that have a part of a ray between l and m are interior angles. Angles 3, 4, 5, 6 are interior angles. • The angles that do not have a part of a ray between l and m are exterior angles. Angles 1, 2, 7, 8 are exterior angles. • Alternate interior angles are on opposite sides of the transversal and do not have a common vertex. Angles 3 and 6 are alternate interior angles, and angles 4 and 5 are alternate interior angles. • Alternate exterior angles are on opposite sides of the transversal and do not have a common vertex. Angles 1 and 8 are alternate exterior angles, and angles 2 and 7 are alternate exterior angles. • Interior angles on the same side of the transversal do not have a common vertex. Angles 3 and 5 are interior angles on the same side of the transversal, and angles 4 and 6 are interior angles on the same side of the transversal. • Corresponding angles are one exterior and one interior angle that are on the same side of the transversal and do not have a common vertex. Angles 1 and 5, angles 2 and 6, angles 3 and 7, and angles 4 and 8 are pairs of corresponding angles. 14365C |
09.pgs 7/10/07 8:48 AM Page 331 In the diagram shown on page 330, the two lines cut by the transversal are not parallel lines. However, when two lines are parallel, many statements may be postulated and proved about these angles. Proving Lines Parallel 331 Theorem 9.1a If two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the two lines are parallel. Given and g CD g AB and F, respectively; 1 2. are cut by transversal Prove g AB g CD g EF at points E A C E 1 2 F B D Proof To prove this theorem, we will use an indirect proof. g AB g AB 1. 2. Statements is not parallel to g. CD and g CD transversal are cut by g EF at points. and F, respectively. g g AB CD point P, forming EFP. and intersect at some 4. m1 m2 5. But 1 2. 6. m1 m2 g AB g CD 7. Reasons 1. Assumption. 2. Given. 3. Two distinct coplanar lines are either parallel or intersecting. 4. The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. 5. Given. 6. Congruent angles are equal in measure. 7. Contradiction in steps 4 and 6. Now that we have proved Theorem 9.1, we can use it in other theorems that also prove that two lines are parallel. Theorem 9.2a If two coplanar lines are cut by a transversal so that the corresponding angles are congruent, then the two lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 332 332 Parallel Lines Given g EF intersects g AB and g CD ; 1 5. Prove g AB g CD Proof Statements g AB intersects 1. g EF 1 5 2. 1 3 3. 3 5 g CD g AB 4 Reasons and g ; CD 1. Given. 2. Vertical angles are congruent. 3. Transitive property of congruence. 4. If two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the two lines are parallel. Theorem 9.3a If two coplanar lines are cut by a trans |
versal so that the interior angles on the same side of the transversal are supplementary, then the lines are parallel. Given intersects g EF of 4. Prove g AB g CD g AB and g CD, and 5 is the supplement E A C 4 3 5 F B D Proof Angle 4 and angle 3 are supplementary since they form a linear pair. If two angles are supplements of the same angle, then they are congruent. Therefore, 3 5. Angles 3 and 5 are a pair of congruent alternate interior angles. If two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the lines are parallel. Therefore, g AB g CD. Theorem 9.4 If two coplanar lines are each perpendicular to the same line, then they are parallel. Given g AB ⊥ g EF and g CD ⊥ g EF. Prove g AB g CD Strategy Show that a pair of alternate interior angles are congruent. A E B 1 C 2 D F 14365C09.pgs 7/10/07 8:48 AM Page 333 Proving Lines Parallel 333 The proof of Theorem 9.4 is left to the student. (See exercise 10.) Methods of Proving Lines Parallel To prove that two coplanar lines that are cut by a transversal are parallel, prove that any one of the following statements is true: 1. A pair of alternate interior angles are congruent. 2. A pair of corresponding angles are congruent. 3. A pair of interior angles on the same side of the transversal are supplementary. 4. Both lines are perpendicular to the same line. EXAMPLE 2 If mA 100 3x and mB 80 3x,. explain why BC AD D C Solution mA mB 100 3x 80 3x 100 80 3x 3x 180 100 3x 80 3x A B Thus, A and B are supplementary. Since AB the segments are parallel, namely, are cut by transversal to form supplementary interior angles on the same side of the transversal, BC and AD AD BC. EXAMPLE 3 If BD bisects ABC, and BC CD, prove CD h BA. C Proof (1) Since BC CD, CBD D because the base B D angles of an isosceles triangle are congruent. (2) Since BD bisects ABC, CBD DBA because the bis |
ector of an angle divides the angle into two congruent angles. A (3) Therefore, by the transitive property of congruence, DBA D. (4) Then, DBA and D are congruent alternate interior angles when h BA are intersected by transversal and two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the two lines are parallel.. Therefore, BD CD because if CD h BA 14365C09.pgs 7/10/07 8:48 AM Page 334 334 Parallel Lines Exercises Writing About Mathematics 1. Two lines are cut by a transversal. If 1 and 2 are vertical angles and 1 and 3 are alternate interior angles, what type of angles do 2 and 3 form? 2. Is it true that if two lines that are not parallel are cut by a transversal, then the alternate interior angles are not congruent? Justify your answer. Developing Skills In 3–8, the figure shows eight angles formed when g AB and g CD are cut by transversal. For each of the following, state the theorem or theorems g EF g AB g. CD || that prove E 12. m3 70 and m5 70 5. m3 60 and m6 120 7. m2 160 and m8 160 4. m2 140 and m6 140 6. m2 150 and m5 30 8. m4 110 and m7 70 Applying Skills 9. Write an indirect proof of Theorem 9.2a, “If two coplanar lines are cut by a transversal so that the corresponding angles are congruent, then the two lines are parallel.” 10. Prove Theorem 9.4, “If two coplanar lines are each perpendicular to the same line, then they are parallel.” In 11 and 12, ABCD is a quadrilateral. 11. If mA 3x and mB 180 3x. Show that AD. BC 12. If DC ⊥ BC and mADC 90, prove AD. BC 13. If bisect each other at AB and CD point E, prove: a. CEA DEB b. ECA EDB c. CA DB C B E A D D A C B 14. Prove that if two coplanar lines are cut by a transversal, forming a pair of alternate exterior angles |
that are congruent, then the two lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 335 Properties of Parallel Lines 335 9-2 PROPERTIES OF PARALLEL LINES In the study of logic, we learned that a conditional and its converse do not always have the same truth value. Once a conditional statement has been proved to be true, it may be possible to prove that its converse is also true. In this section, we will prove converse statements of some of the theorems proved in the previous section. The proof of these converse statements requires the following postulate and theorem: Postulate 9.2 Through a given point not on a given line, there exists one and only one line parallel to the given line. Theorem 9.5 If, in a plane, a line intersects one of two parallel lines, it intersects the other. Given g AB g CD and g EF intersects g AB at H. Prove g EF intersects g. CD Proof Assume g g EF CD g EF does not intersect g CD. Then. Therefore, through H, a point g AB, two lines, g EF and are g CD not on E A H F C B D g CD each parallel to postulate that states that through a given point not on a given line, one and only one line can be drawn parallel to a given line. Since our assumption leads. This contradicts the to a contradiction, the assumption must be false and its negation, sects g CD must be true. g EF inter- Now we are ready to prove the converse of Theorem 9.1a. Theorem 9.1b If two parallel lines are cut by a transversal, then the alternate interior angles formed are congruent. Given g AB g CD g CD at F., transversal g EF intersects g AB at E and Prove 14365C09.pgs 7/10/07 8:48 AM Page 336 336 Parallel Lines Proof We can use an indirect proof. Assume 1 is so that not congruent to 2. Construct HEF 2. Since HEF and 2 are congruent alternate interior angles, h EH g HE g CD. But g line through E, and we are given AB contradicts the postulate that states that through a given point not on a given line, there exists one and only one line parallel to the given line. Thus, the |
assumption is false and 1 2.. This is a D C F g AB g CD 2 H A E B Note that Theorem 9.1b is the converse of Theorem 9.1a. We may state the two theorems in biconditional form: Theorem 9.1 Two coplanar lines cut by a transversal are parallel if and only if the alternate interior angles formed are congruent. Each of the next two theorems is also a converse of a theorem stated in Section 9-1. Theorem 9.2b If two parallel lines are cut by a transversal, then the corresponding angles are congruent. (Converse of Theorem 9.2a) Given g AB g CD and transversal g EF Prove Proof Statements g 1. CD 2. 3 5 g AB and transversal 3. 1 3 4. 1 5 Reasons g EF 1. Given. 2. If two parallel lines are cut by a transversal, then the alternate interior angles formed are congruent. 3. Vertical angles are congruent. 4. Transitive property of congruence. 14365C09.pgs 7/10/07 8:48 AM Page 337 Theorem 9.3b If two parallel lines are cut by a transversal, then two interior angles on the same side of the transversal are supplementary. (Converse of Theorem 9.3a) Properties of Parallel Lines 337 Given g AB g CD and transversal g EF Prove 4 is the supplement of 5. Strategy Show that 3 5 and that 4 is the supplement of 3. If two angles are congruent, then their supplements are congruent. Therefore, 4 is also the supplement of 5. E 34 B D A C 5 F The proof of this theorem is left to the student. (See exercise 18.) Since Theorems 9.2b and 9.3b are converses of Theorems 9.2a and 9.3a, we may state the theorems in biconditional form: Theorem 9.2 Two coplanar lines cut by a transversal are parallel if and only if corresponding angles are congruent. Theorem 9.3 Two coplanar lines cut by a transversal are parallel if and only if interior angles on the same side of the transversal are supplementary. EXAMPLE 1 Transversal g EF intersects |
g AB g CD and g AB g at G and H, CD, mBGH 3x 20, and respectively. If mGHC 2x 10: a. Find the value of x. c. Find mGHD. b. Find mGHC. A C E G B D H F Solution a. Since g AB g CD and these lines are cut by transversal g EF, the alternate inte- rior angles are congruent: mBGH mGHC 3x 20 2x 10 3x 2x 10 20 x 30 b. mGHC 2x 10 2(30) 10 70 14365C09.pgs 7/10/07 8:48 AM Page 338 338 Parallel Lines c. Since GHC and GHD form a linear pair and are supplementary, mGHD 180 mGHC 180 70 110 Answers a. x 70 b. mGHC 70 c. mGHD 110 Using Theorem 9.1, we may also prove the following theorems: Theorem 9.6 If a transversal is perpendicular to one of two parallel lines, it is perpendicular to the other. Given g AB g CD, g EF ⊥ g AB Prove g EF ⊥ g CD A C E B F D Strategy Show that alternate interior angles are right angles. The proof of this theorem is left to the student. (See exercise 19.) Theorem 9.7 If two of three lines in the same plane are each parallel to the third line, then they are parallel to each other. Given g AB g LM and g CD g LM Prove g AB g CD Proof Draw transversal at H. Since g AB intersecting g LM A, this transversal also intersects Similarly, since sal also intersects. Call this point F. g LM g CD at a point G., this transver- g EJ g LM g AB g CD AB Since g LM, alternate interior angles formed are congruent. Therefore, AFG GHM. Similarly, since, CGH GHM. By the transitive property of congruence, AFG CGH. Angles AFG and CGH are conare intersected by transversal gruent corresponding angles when g LM g CD g CD g AB and 14365C09.pgs 7/10/07 8:48 AM Page 339 Properties of Parallel Lines 339 g EJ. Therefore, g AB g CD because if two coplanar lines |
are cut by a transversal so that the corresponding angles formed are congruent, then the two lines are parallel. SUMMARY OF PROPERTIES OF PARALLEL LINES If two lines are parallel: 1. A transversal forms congruent alternate interior angles. 2. A transversal forms congruent corresponding angles. 3. A transversal forms supplementary interior angles on the same side of the transversal. 4. A transversal perpendicular to one line is also perpendicular to the other. 5. A third line in the same plane that is parallel to one of the lines is parallel to the other. EXAMPLE 2 Given: Quadrilateral ABCD, BC DA, and BC DA B Prove: AB CD Proof Use congruent triangles to prove congruent alternate A C D interior angles. Statements 1. BC DA C BC 2. DA 3. BCA DAC AC 4. AC 5. BAC DCA 6. BAC DCA C 7. AB CD A A B B D D Reasons 1. Given. 2. Given. 3. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 4. Reflexive property of congruence. 5. SAS. 6. Corresponding parts of congruent triangles are congruent. 7. If two lines cut by a transversal form congruent alternate interior angles, the lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 340 340 Parallel Lines Note: In the diagram for Example 2, you may have noticed that two parallel lines, g BC and g DA, each contained a single arrowhead in the same direction. Such pairs of arrowheads are used on diagrams to indicate that two lines are parallel. Exercises Writing About Mathematics 1. a. Is the inverse of Theorem 9.1a always true? Explain why or why not. b. Is the inverse of Theorem 9.6 always true? Explain why or why not. 2. Two parallel lines are cut by a transversal forming alternate interior angles that are supplementary. What conclusion can you draw about the measures of the angles formed by the parallel lines and the transversal. Justify your answer. Developing Skills g AB In 3–12, g EF are cut by transversal g CD 3. m5 when m3 80. 5. m4 when m5 60. 7. |
m8 when m3 65. 9. m3 when m3 3x and m5 x 28. 10. m5 when m3 x and m4 x 20. 11. m7 when m1 x 40 and m2 5x 10. 12. m5 when m2 7x 20 and m8 x 100. as shown in the diagram. Find: 4. m2 when m6 150. 6. m7 when m1 75. 8. m5 when m2 130. 12 13. Two parallel lines are cut by a transversal. For each pair of interior angles on the same side of the transversal, the measure of one angle exceeds the measure of twice the other by 48 degrees. Find the measures of one pair of interior angles. 14. Two parallel lines are cut by a transversal. The measure of one of the angles of a pair of corresponding angles can be represented by 42 less than three times the other. Find the measures of the angles of this pair of corresponding angles. 15. In the diagram, and a. If mFGD 110 and mFEC 130, find the measures intersect and at F. g AFB g CD g EF g GF g AB of each of the angles numbered 1 through 9. b. What is the measure of an exterior angle of EFG at F 14365C09.pgs 7/10/07 8:48 AM Page 341 Properties of Parallel Lines 341 c. Is the measure of an exterior angle at F greater than the measure of either of the non- adjacent interior angles? d. What is the sum of the measures of the nonadjacent interior angles of an exterior angle at F? e. What is the sum of the measures of the nonadjacent interior angles of the exterior angle, FGD? f. What is the sum of the measures of the nonadjacent interior angles of the exterior angle, FEC? g. What is the sum of the measures of the angles of EFG? 16. Two pairs of parallel lines are drawn; g ABE g DC and g AD g BC. If mCBE 75, find the measure of each angle of quadrilateral ABCD. Applying Skills D C A B E 17. Prove Theorem 9.3b, “If two parallel lines are cut by a transversal, then two interior angles on the same side of the transversal are supplementary.” 18. Prove Theorem 9.6, “If a |
transversal is perpendicular to one of two parallel lines, it is per- pendicular to the other.” 19. Prove that if two parallel lines are cut by a transversal, the alternate exterior angles are con- gruent. 20. Given: ABC, h CE Prove: A B. bisects exterior BCD, and h CE AB. 21. Given: CAB DCA and DCA ECB Prove: a. g AB DCE. b. CAB is the supplement of CBG. 22. The opposite sides of quadrilateral PQRS are parallel, that is, is a right angle, prove that Q, R, and S are right angles PQ RS and QR SP. If P 23. The opposite sides of quadrilateral KLMN are parallel, that is, LM NK. If K is an acute angle, prove that M is an acute angle and that L and N are obtuse angles. KL MN and 14365C09.pgs 7/10/07 8:48 AM Page 342 342 Parallel Lines 9-3 PARALLEL LINES IN THE COORDINATE PLANE In Chapter 6 we stated postulates about horizontal and vertical lines in the coordinate plane. One of these postulates states that each vertical line is perpendicular to each horizontal line. We can use this postulate to prove the following theorem: Theorem 9.8 If two lines are vertical lines, then they are parallel. Proof: Since each vertical line is perpendicular to each horizontal line, each vertical line is perpendicular to the x-axis, a horizontal line. Theorem 9.6 states that if two coplanar lines are each perpendicular to the same line, then they are parallel. Therefore, all vertical lines are parallel. A similar theorem can be proved about horizontal lines: Theorem 9.9 If two lines are horizontal lines, then they are parallel. Proof: Since each horizontal line is perpendicular to each vertical line, each horizontal line is perpendicular to the y-axis, a vertical line. Theorem 9.6 states that if two coplanar lines are each perpendicular to the same line, then they are parallel. Therefore, all horizontal lines are parallel. We know that all horizontal lines have the same slope, 0. We also know that all vertical lines have no slope. Do parallel lines that are neither horizontal nor vertical have the same slope? When we draw parallel lines in the coordinate plane, it appears that this is true |
. Theorem 9.10a If two non-vertical lines in the same plane are parallel, then they have the same slope. Given l1 l2 Prove The slope of l1 is equal to slope of l2. Proof In the coordinate plane, let the slope of l1 be m 0. Choose any point on l1. Through a given point, one and only one line can be drawn perpendicular to a given line. Through that point, draw k, a line perpendicular to l1. y k O l1 l2 x 14365C09.pgs 7/10/07 8:48 AM Page 343 Parallel Lines in the Coordinate Plane 343 If two lines are perpendicular, the slope of one is the negative reciprocal l2 of the slope of the other. Therefore, the slope of k is. It is given that. l1 Then, k is perpendicular to l2 because if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other. The slope of l2 is the negative reciprocal of the slope of k. The negative reciprocal of is m. Therefore, the slope of l1 is equal to the slope of l2. 21 m 21 m Is the converse of this statement true? We will again use the fact that two lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other to prove that it is. Theorem 9.10b If the slopes of two non-vertical lines in the coordinate plane are equal, then the lines are parallel. k y O l1 l2 x Given Lines l1 and l2 with slope m Prove l1 l2 Proof Choose any point on l1. Through a given point, one and only one line can be drawn perpendicular to a given line. Through that point, draw k, a line perpendicular to l1. The slope of k is 21 since two non-vertical lines are m perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other. But this means that l2 Therefore, l1 parallel. ⊥ k because the slope of l2 is also the negative reciprocal of the slope of k. l2 because two lines perpendicular to the same line are We can write the statements that we have proved as a biconditional: Theorem 9.10 Two non-vertical lines in the coordinate plane are parallel if and only if they have the same slope. EXAMPLE 1 |
The vertices of quadrilateral ABCD are A(2, 4), B(6, 2), C(2, 6), and D(1, 2). a. Show that two sides of the quadrilateral are parallel. b. Show that the quadrilateral has two right angles. 14365C09.pgs 7/10/07 8:48 AM Page 344 344 Parallel Lines Solution The slope of AB 5 22 2 (2422) 2 2 6 5 8 21 2 2 5 24 2 2 (21) 5 26 BC 5 CD 5 2 2 6 DA 5 24 2 2 24 5 22 23 5 4. 3 3 5 22.. The slope of The slope of The slope of a. and BC DA equal slopes. are parallel because they have b. The slope of AB the slope of BC Therefore, B is a right angle. is the negative reciprocal of, so they are perpendicular The slope of perpendicular. Therefore, A is a right angle. is the negative reciprocal of the slope of AB DA, so they are Answers a. DABC b. A and B are right angles. EXAMPLE 2 Write an equation for l1, the line through (2, 5) that is parallel to the line l2 whose equation is 2x y 7. Solution (1) Solve the equation of l2 for y: (2) Find the slope of l2. The slope of a line in slope-intercept form is the coefficient of x: (3) Find the slope of l1, which is equal to the slope of l2: (4) Use the definition of slope to write an equation of l1. Let (x, y) and (2, 5) be two points on l1: Answer y 2x 1 or 2x y 1 2x y 7 y 2x 7 y 2x 7 slope slope of l1 2 slope y1 2 y2 x1 2 x2 y 2 5 x 2 (22) 2 y 5 2(x 2) y 5 –2x 4 y 2x 1 14365C09.pgs 7/10/07 8:48 AM Page 345 Parallel Lines in the Coordinate Plane 345 Exercises Writing About Mathematics 1. If l1 and l2 have the same slope and have a common point, what must be true about l1 and l2? 2. Theorem 9.10 is true for all lines that are not vertical. Do vertical lines have the same slope? Explain |
your answer. Developing Skills In 3–8, for each pair of lines whose equations are given, tell whether the lines are parallel, perpendicular, or neither parallel nor perpendicular. 3. x y = 7 x y 3 5. x 1 3y 1 2 y 3x 2 7. x = 2 x 5 4. 2x y 5 y 2x 3 6. 2x y 6 2x y = 3 8. x = 2 y 3 In 9–12, write an equation of the line that satisfies the given conditions. 9. Parallel to y 3x 1 with y-intercept 4. 10. Perpendicular to y 3x 1 with y-intercept 4. 11. Parallel to x 2y 4 and through the point (4, 5). 12. Parallel to and 3 units below the x-axis. Applying Skills 13. Quadrilateral ABCD has two pairs of parallel sides,. The coordinates of A are (1, 2), the coordinates of B are (7, 1) and the coordinates of C are (8, 2). and AB CD BC DA a. What is the slope of? AB b. What is the slope of c. Write an equation for d. What is the slope of? BC e. What is the slope of f. Write an equation for? CD g. CD? AD g. AD g. Use the equation of g CD and the equation of g AD to find the coordinates of D. 14365C09.pgs 7/10/07 8:48 AM Page 346 346 Parallel Lines 14. In quadrilateral ABCD, BC'AB, DA'AB, and DA'DC. The coordinates of A are (1, 1), the coordinates of B are (4, 2), and the coordinates of C are (2, 4). a. What is the slope of b. What is the slope of c. What is the slope of d. What is the slope of? AB? BC? Justify your answer. AD? Justify your answer. DC e. Write an equation for f. Write an equation for g. DC g. AD g. Use the equation of g DC and the equation of g AD to find the coordinates of D. 15. The coordinates of the vertices of quadrilateral PQRS are P(0, 1), Q(4, 0), R(2, 3), and S(2, 2). a. Show that PQRS has two pairs of |
parallel sides. b. Show that PQRS does not have a right angle. 16. The coordinates of the vertices of quadrilateral KLMN are K(2, 1), L(4, 3), M(2, 1), and N(1, 2). a. Show that KLMN has only one pair of parallel sides. b. Show that KLMN has two right angles. Hands-On Activity 1 In this activity, we will use a compass and a straightedge, or geometry software to construct a line parallel to a given line through a point not on the line. STEP 1. Given a point, P, not on line, l. Through P, construct a line perpen- dicular to line l. Label this line n. P l STEP 2. Through P, construct a line, p, perpendicular to line n. Result: l p a. Justify the construction given in the procedure. b. In (1)–(3), construct a line parallel to the line through the given point. (1) y 1 4x 1 5 ; 3, 51 2 B A (2) 12x y 19; (12, 4) (3) y 21 9x 2 3 ; (0, 4) Hands-On Activity 2 A midsegment is a segment formed by joining two midpoints of the sides of a triangle. In this activity, we will prove that a midsegment is parallel to the third side of the triangle using coordinate geometry. 1. With a partner or in a small group, complete the following: a. Write the coordinates of a triangle using variables. These coordinates can be any convenient variables. midsegment 14365C09.pgs 7/10/07 8:48 AM Page 347 The Sum of the Measures of the Angles of a Triangle 347 b. Find the midpoints of two sides of the triangle. c. Prove that the midsegment formed is parallel to the third side of the triangle. 2. Compare your proof with the proofs of the other groups. Were different coordinates used? Which coordinates seem easiest to work with? 9-4 THE SUM OF THE MEASURES OF THE ANGLES OF A TRIANGLE In previous courses, you have demonstrated that the sum of the measures of the angles of a triangle is 180 degrees. The congruent angles formed when parallel lines are cut by a transversal make it possible for us to prove this fact. Theorem 9.11 The |
sum of the measures of the angles of a triangle is 180°. Given ABC Prove mA mB mC 180 D A B E C Proof Statements Reasons 1. Let g DE be the line through B that is parallel to. AC 2. mDBE 180 3. mDBA mABC mCBE 180 4. A DBA and C CBE 5. mA mDBA and mC CBE 1. Through a given point not on a given line, there exists one and only one line parallel to the given line. 2. A straight angle is an angle whose degree measure is 180. 3. The whole is equal to the sum of all its parts. 4. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 5. Congruent angles are equal in measure. 6. mA mABC mC 180 6. Substitution postulate. Many corollaries to this important theorem exist. 14365C09.pgs 7/10/07 8:48 AM Page 348 348 Parallel Lines Corollary 9.11a If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent. Proof: Let ABC and DEF be two triangles in which A D and B E. Since the sum of the degree measures of the angles of a triangle is 180, then mA mB mC mD mE mF. We use the subtraction postulate to prove that mC mF and therefore, that C F. Corollary 9.11b The acute angles of a right triangle are complementary. Proof: In any triangle ABC, mA mB mC 180. If C is a right angle, mC 90, mA mB 90 180 mA mB 90 Therefore, A and B are complementary. Corollary 9.11c Each acute angle of an isosceles right triangle measures 45°. Proof: In isosceles right triangle ABC, mC 90 and. Therefore, mA = mB. Using Corollary 9.12b, we know that A and B are complementary. Therefore, the measure of each must be 45. AC BC Corollary 9.11d Each angle of an equilateral triangle measures 60°. Proof: In equilateral triangle ABC, mA mB mC. We substitute mA for mB and mC in the equation mA mB |
mC 180, and then solve the resulting equation: 3mA 180 so mA 60. Corollary 9.11e The sum of the measures of the angles of a quadrilateral is 360°. Proof: In quadrilateral ABCD, we draw, forming two triangles. The sum of AC the measures of the angles of quadrilateral ABCD is the sum of the measures of the angles of the two triangles: D C 180 180 360 A B 14365C09.pgs 7/10/07 8:48 AM Page 349 The Sum of the Measures of the Angles of a Triangle 349 Corollary 9.11f The measure of an exterior angle of a triangle is equal to the sum of the measures of the nonadjacent interior angles. The proof is left to the student. (See exercise 30.) Note: Recall that the Exterior Angle Theorem of Section 7-5 gives an inequality that relates the exterior angle of a triangle to the nonadjacent interior angles: “The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle.” Corollary 9.11f is a version of the Exterior Angle Theorem involving equality. EXAMPLE 1 The measure of the vertex angle of an isosceles triangle exceeds the measure of each base angle by 30 degrees. Find the degree measure of each angle of the triangle. Solution Let x measure of each base angle. Let x 30 measure of vertex angle. The sum of the measures of the angles of a triangle is 180. x x x 30 180 3x 30 180 3x 150 x 50 x 30 80 Answer The angle measures are 50°, 50°, and 80°. EXAMPLE 2 In ABC, the measures of the three angles are represented by 9x, 3x 6, and 11x 2. Show that ABC is a right triangle. Solution Triangle ABC will be a right triangle if one of the angles is a right angle. Write and equation for the sum of the measures of the angles of ABC. 9x 3x 6 11x 2 180 23x 4 180 23x 184 x 8 14365C09.pgs 7/10/07 8:48 AM Page 350 350 Parallel Lines Substitute x 8 in the representations of the angle measures. 9x 9(8) 72 3x 6 3(8) 6 24 6 18 11x 2 11(8) 2 88 2 90 Answer Triangle ABC is a right triangle because the |
degree measure of one of its angles is 90. EXAMPLE 3 B is a not a point on DCB and AC > BC g ACD. Ray h CE. Prove that bisects g CE. g AB Solution Given: h CE bisects DCB and AC > BC. Prove: g AB g CE Proof Statements h CE bisects DCB. 1. 2. DCE ECB 3. mDCE mECB AC > BC 4. 5. mCAB mCBA 6. mDCB mCAB mCBA 7. mDCB mDCE mECB 8. mDCE mECB mCAB mCBA 9. mDCE mDCE mCAB mCAB or 2mDCE 2 mCAB 10. mDCE mCAB g CE g AB 11. B C E D A Reasons 1. Given. 2. Definition of an angle bisector. 3. Measures of congruent angles are equal. 4. Given. 5. Isosceles triangle theorem. 6. An exterior angle of a triangle is equal to the sum of the measures of the nonadjacent interior angles. 7. Partition postulate. 8. Substitution postulate (steps 6 and 7). 9. Substitution postulate (steps 3, 5, and 8). 10. Division postulate. 11. If two lines are cut by a transversal forming equal corresponding angles, then the lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 351 The Sum of the Measures of the Angles of a Triangle 351 Exercises Writing About Mathematics 1. McKenzie said that if a triangle is obtuse, two of the angles of the triangle are acute. Do you agree with McKenzie? Explain why or why not. 2. Giovanni said that since the sum of the measures of the angles of a triangle is 180, the angles of a triangle are supplementary. Do you agree with Giovanni? Explain why or why not. Developing Skills In 3–6, determine whether the given numbers can be the degree measures of the angles of a triangle. 3. 25, 100, 55 4. 95, 40, 45 5. 75, 75, 40 6. 12, 94, 74 In 7–10, the given numbers are the degree measures of two angles of a triangle. Find the measure of the third angle. 7. 80, 60 8. 45, 85 |
9. 90, 36 10. 65, 65 In 11–14, the measure of the vertex angle of an isosceles triangle is given. Find the measure of a base angle. 11. 20 12. 90 13. 76 14. 110 In 15–18, the measure of a base angle of an isosceles triangle is given. Find the measure of the vertex angle. 15. 80 16. 20 17. 45 18. 63 19. What is the measure of each exterior angle of an equilateral triangle? In 20–23, the diagram shows ABC and exterior ACD. 20. If mA 40 and mB 20, find mACD and mACB. 21. If mA 40 and mB 50, find mACD and mACB. 22. If mA 40 and mACB 120, find mACD and mB. 23. If mA 40, mB 3x 20, and mACD 5x 10, find mB, mACD, and mACB. Applying Skills D C A B 24. The measure of each base angle of an isosceles triangle is 21° more than the measure of the vertex angle. Find the measure of each angle of the triangle. 14365C09.pgs 7/10/07 8:48 AM Page 352 352 Parallel Lines 25. The measure of an exterior angle at C of isosceles ABC is 110°. If AC = BC, find the mea- sure of each angle of the triangle. 26. The measure of an exterior angle at D of isosceles DEF is 100°. If DE = EF, find the mea- sure of each angle of the triangle. 27. Triangle LMN is a right triangle with M the right angle. If mL 32, find the measure of N and the measure of the exterior angle at N. 28. In ABC, mA 2x 18, mB x 40, and mC 3x – 40. a. Find the measure of each angle of the triangle. b. Which is the longest side of the triangle? 29. The measure of an exterior angle at B, the vertex of isosceles ABC, can be represented by 3x 12. If the measure of a base angle is 2x 2, find the measure of the exterior angle and of the interior angles of ABC. 30. Prove Corollary 9.11f, “The measure of an |
exterior angle of a triangle is equal to the sum of the measures of the nonadjacent interior angles.” 31. a. In the coordinate plane, graph points A(5, 2), B(2, 2), C(2, 1), D(1, 1). g AB and BDC. b. Draw c. Explain how you know that BDC is an isosceles right triangle. d. What is the measure of BDC? Justify your answer. e. What is the measure of DBA? Justify your answer. 32. Prove that the sum of the measures of the angles of hexagon ABCDEF is 720°. (Hint: draw AD.) 33. ABCD is a quadrilateral with h BD the bisector of ABC and h DB the bisector of ADC. Prove that A C. 9-5 PROVING TRIANGLES CONGRUENT BY ANGLE, ANGLE, SIDE When two angles of one triangle are congruent to two angles of another triangle, the third angles are congruent. This is not enough to prove that the two triangles are congruent. We must know that at least one pair of corresponding sides are congruent. We already know that if two angles and the side between them in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent by ASA. Now we want to prove angle-angle-side or AAS triangle congruence. This would allow us to conclude that if any two angles and any side in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent. 14365C09.pgs 7/10/07 8:48 AM Page 353 Proving Triangles Congruent by Angle, Angle, Side 353 Theorem 9.12 If two angles and the side opposite one of them in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent. (AAS) Given ABC and DEF, A D, C F, and AB > DE Prove ABC DEF C F A B D E Proof Statements Reasons 1. A D 2. C F 3. B E 1. Given. 2. Given. 3. If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent. AB > DE 4. 5. ABC DEF |
4. Given. 5. ASA. Therefore, when two angles and any side in one triangle are congruent to the corresponding two angles and side of a second triangle, we may say that the triangles are congruent either by ASA or by AAS. The following corollaries can proved using AAS. Note that in every right tri- angle, the hypotenuse is the side opposite the right angle. Corollary 9.12a Two right triangles are congruent if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of the other right triangle. The proof uses AAS and is left to the student. (See exercise 15.) Corollary 9.12b If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. Recall that the distance from a point to a line is the length of the perpendicular from the point to the line. The proof uses AAS and is left to the student. (See exercise 16.) You now have four ways to prove two triangles congruent: SAS, ASA, SSS, and AAS. A P C B 14365C09.pgs 7/10/07 8:48 AM Page 354 354 Parallel Lines EXAMPLE 1 Prove that the altitudes drawn to the legs of an isosceles triangle from the endpoints of the base are congruent. B Given: Isosceles triangle ABC with BA, BC AE ⊥ BC, D E and CD ⊥ BA. Prove: CD AE A C Proof Statements Reasons 1. In ABC, BA. BC 1. Given. E C C A 2. BAC BCA ⊥ AE, BC 3. CD 4. CDA and AEC are right BA ⊥ angles. A 5. CDA AEC AC 7. DAC ECA S 6. AC 8. CD AE 2. If two sides of a triangle are congruent, the angles opposite these sides are congruent. 3. Given. 4. Perpendicular lines are two lines that intersect to form right angles. 5. All right angles are congruent. 6. Reflexive property of congruence. 7. AAS (steps 2, 5, and 6). 8. Corresponding parts of congruent triangles are congruent. B B A A D EXAMPLE 2 The coordinates |
of the vertices of ABC are A(6, 0), B(1, 0) and C(5, 2). The coordinates of DEF are D(3, 0), E(8, 0), and F(4, 2). Prove that the triangles are congruent. C(5, 2) y F(4, 2) B(1, 0) O A(6, 0) D(3, 0) E(8, 0) x Solution (1) Prove that the triangles are right triangles. In ABC: 2 2 0 The slope of AC is The slope of CB is 25 2 (26) 5 2 25 2 (21) 5 2 2 2 0 1 5 2. 24 5 21 2 The slope of. The slope of In DEF: DF is 2 2 0 FE is. 24 5 21 2. Two lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. Therefore, ACB is a right triangle. Also, DFE is a right triangle., DFE is a right angle, and, ACB is a right angle, and DF'FE AC'CB 14365C09.pgs 7/10/07 8:48 AM Page 355 Proving Triangles Congruent by Angle, Angle, Side 355 (2) Prove that two acute angles are congruent. Two lines are parallel if their slopes are equal. Therefore,. The x-axis is a transversal forming congruent corresponding angles, so CBA and FED are congruent. CB FE (3) Prove that the hypotenuses are congruent. The hypotenuse of ABC is hypotenuse of DEF is DE have the same measure are congruent, and so, and AB |–6 (1)| 5. The AB, and DE |3 8| 5. Line segments that AB > DE. (4) Therefore, ABC DEF because the hypotenuse and an acute angle of one triangle are congruent to the hypotenuse and an acute angle of the other. EXAMPLE 3 Show that if a triangle has two sides and an angle opposite one of the sides congruent to the corresponding sides and angle of another triangle, the triangles may not be congruent. Solution (1) Draw an angle, ABC. A (2) Open a compass to a length that is smaller than AB but larger than the distance from A to. Use the compass to |
mark two points, D and E, on BC BC. (3) Draw AD (4) In ABD and ABE, and AE. AB AD > AE, AB. In these B D E C A B B, and two triangles, two sides and the angle opposite one of the sides are congruent to the corresponding parts of the other triangle. But ABD and ABE are not congruent. This counterexample proves that SSA is not sufficient to prove triangles congruent. D E C B Note: Triangles in which two sides and an angle opposite one of them are congruent may not be congruent to each other. Therefore, SSA is not a valid proof of triangle congruence. Similarly, triangles in which all three angles are congruent may not be congruent to each other, so AAA is also not a valid proof of triangle congruence. 14365C09.pgs 7/10/07 8:48 AM Page 356 356 Parallel Lines Exercises Writing About Mathematics 1. In Example 3, we showed that SSA cannot be used to prove two triangles congruent. Does this mean that whenever two sides and an angle opposite one of the sides are congruent to the corresponding parts of another triangle the two triangles are not congruent? Explain your answer. 2. In the coordinate plane, points A and C are on the same horizontal line and C and B are on the same vertical line. Are CAB and CBA complementary angles? Justify your answer. Developing Skills In 3–8, each figure shows two triangles. Congruent parts of the triangles have been marked. Tell whether or not the given congruent parts are sufficient to prove that the triangles are congruent. Give a reason for your answer. 3. D C 4. D A B A E C B 5. F E 6. C 7. B C 8. A D A BD Applying Skills A R B Q S P 9. Prove that if two triangles are congruent, then the altitudes drawn from corresponding ver- tices are congruent. 10. Prove that if two triangles are congruent, then the medians drawn from corresponding ver- tices are congruent. 11. Prove that if two triangles are congruent, then the angle bisectors drawn from correspond- ing vertices are congruent. 14365C09.pgs 7/10/07 |
8:48 AM Page 357 The Converse of the Isosceles Triangle Theorem 357 12. Given: Quadrilateral ABCD with A C and h BD the bisector of ABC. h DB bisects ADC. Prove: 13. Given: AB CD, AB > CD, and AB'BEC. A Prove: AED and BEC bisect each other. B E C C A B D D 14. a. Use a translation to prove that ABC and DEF in Example 2 are congruent. b. Use two line reflections to prove that ABC and DEF in Example 2 are congruent. 15. Prove Corollary 9.12a, “Two right triangles are congruent if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of the other right triangle.” 16. Prove Corollary 9.12b, “If a point lies on the bisector of an angle, it is equidistant from the sides of the angle.” 17. Prove that if three angles of one triangle are congruent to the corresponding angles of another (AAA), the triangles may not be congruent. (Through any point on side of ABC, draw a line segment parallel to AC BC.) 9-6 THE CONVERSE OF THE ISOSCELES TRIANGLE THEOREM The Isosceles Triangle Theorem, proved in Section 5-3 of this book, is restated here in its conditional form. If two sides of a triangle are congruent, then the angles opposite these sides are congruent. When we proved the Isosceles Triangle Theorem, its converse would have been very difficult to prove with the postulates and theorems that we had available at that time. Now that we can prove two triangles congruent by AAS, its converse is relatively easy to prove. Theorem 9.13 If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 14365C09.pgs 7/10/07 8:48 AM Page 358 358 Parallel Lines Given ABC with A B. Prove CA CB Proof We can use either the angle bisector or the altitude from C to separate the triangle into two congruent triangles. We will use the angle bisector. A Statements Reasons C D B 1. Draw ACB. CD |
, the bisector of 2. ACD BCD 3. A B CD 4. CD 5. ACD BCD 6. CA CB 1. Every angle has one and only one bisector. 2. An angle bisector of a triangle is a line segment that bisects an angle of the triangle. 3. Given. 4. Reflexive property of congruence. 5. AAS. 6. Corresponding parts of congruent triangles are congruent. The statement of the Isosceles Triangle Theorem (Theorem 5.1) and its con- verse (Theorem 9.14) can now be written in biconditional form: Two angles of a triangle are congruent if and only if the sides opposite these angles are congruent. To prove that a triangle is isosceles, we may now prove that either of the fol- lowing two statements is true: 1. Two sides of the triangle are congruent. 2. Two angles of the triangle are congruent. Corollary 9.13a If a triangle is equiangular, then it is equilateral. Given ABC with A B C. Prove ABC is equilateral. 14365C09.pgs 7/10/07 8:48 AM Page 359 The Converse of the Isosceles Triangle Theorem 359 Proof We are given equiangular ABC. Then since A B, the sides opposite these angles are congruent, that is, AC > AB erty of congruence, AB > BC > CA for the same reason. Therefore, BC > AC. Also, since B C, AC > AB by the transitive prop-, and ABC is equilateral. EXAMPLE 1 In PQR, Q R. If PQ 6x 7 and PR 3x 11, find: a. the value of x b. PQ c. PR Solution a. Since two angles of PQR b. PQ 5 6x 2 7 c. PR 5 3x 1 11 are congruent, the sides opposite these angles are congruent. Thus, PQ PR. 6x 7 3x 11 6x 3x 11 7 3x 18 x 6 Answer 5 6(6) 2 7 5 36 2 7 5 29 Answer 5 3(6) 1 11 5 18 1 11 5 29 Answer EXAMPLE 2 The degree measures of the three angles of ABC are represented by mA x 30, mB |
3x, and mC 4x 30. Describe the triangle as acute, right, or obtuse, and as scalene, isosceles, or equilateral. Solution The sum of the degree measures of the angles of a triangle is 180. x 30 3x 4x 30 180 8x 60 180 8x 120 x 15 Substitute x 15 in the representations given for the three angle measures. m/A 5 x 1 30 5 15 1 30 5 45 m/B 5 3x m/C 5 4x 1 30 5 3(15) 5 45 5 4(15) 1 30 5 60 1 30 5 90 Since A and B each measure 45°, the triangle has two congruent angles and therefore two congruent sides. The triangle is isosceles. Also, since one angle measures 90°, the triangle is a right triangle. Answer ABC is an isosceles right triangle. 14365C09.pgs 7/10/07 8:48 AM Page 360 360 Parallel Lines EXAMPLE 3 Given: Quadrilateral ABCD with and h AC bisects DAB. AB CD D C Prove: AD > CD A B Proof Statements Reasons D C AB CD 1. 2. DCA CAB A B h AC bisects DAB. 3. 4. CAB DAC 5. DCA DAC congruence. 6. AD > CD Exercises Writing About Mathematics 1. Given. 2. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 3. Given. 4. A bisector of an angle divides the angle into two congruent parts. 5. Transitive property of 6. If two angles of a triangle are congruent, the sides opposite these angles are congruent. 1. Julian said that the converse of the Isosceles Triangle Theorem could have been proved as a corollary to Theorem 7.3, “If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal.” Do you agree with Julian? Explain why or why not. 2. Rosa said that if the measure of one angle of a right triangle is 45 degrees, then the triangle is an isosceles right triangle. Do you agree with Rosa? Explain why or why not. 14365C09.pgs 7/10/07 8:48 AM Page 361 The |
Converse of the Isosceles Triangle Theorem 361 In 3–6, in each case the degree measures of two angles of a triangle are given. a. Find the degree measure of the third angle of the triangle. b. Tell whether the triangle is isosceles or is not isosceles. 3. 70, 40 4. 30, 120 6. 80, 40 5. 50, 65 7. In ABC, mA mC, AB 5x 6, and BC 3x 14. Find the value of x. 8. In PQR, mQ mP, PR 3x, and RQ 2x 7. Find PR and RQ. 9. In MNR, MN NR, mM = 72, and mR 2x. Find the measures of R and of N. 10. In ABC, mA 80 and mB 50. If AB 4x 4, AC 2x 16, and BC 4x 6, find the measure of each side of the triangle. 11. The degree measures of the angles of ABC are represented by x 10, 2x, and 2x 30. Show that ABC is an isosceles triangle. 12. The degree measures of the angles of ABC are represented by x 35, 2x 10, and 3x 15. Show that ABC is an equilateral triangle. 13. The degree measures of the angles of ABC are represented by 3x 18, 4x 9, and 10x. Show that ABC is an isosceles right triangle. 14. What is the measure of each exterior angle of an equilateral triangle? 15. What is the sum of the measures of the exterior angles of any triangle? Applying Skills 16. Given: P is not on › ‹ ABCD and ABP PCD. Prove: BPC is isosceles. P A B C D 17. Given: P is not on AB and PAB PBA. Prove: P is on the perpendicular bisector of AB. P A B 14365C09.pgs 7/10/07 8:48 AM Page 362 362 Parallel Lines 18. Given: h BE angle of ABC, and bisects DBC, an exterior AC. h BE Prove: AB > CB E C 19. Given: P is not on, ABCD PBC PCB, and APB DPC Prove: AP > DP P A B D A B C D 20. Prove Theorem |
9.13 by drawing the altitude from C. 9-7 PROVING RIGHT TRIANGLES CONGRUENT BY HYPOTENUSE, LEG We showed in Section 5 of this chapter that, when two sides and an angle opposite one of these sides in one triangle are congruent to the corresponding two sides and angle in another triangle, the two triangles may or may not be congruent. When the triangles are right triangles, however, it is possible to prove that they are congruent. The congruent angles are the right angles, and each right angle is opposite the hypotenuse of the triangle. Theorem 9.14 If the hypotenuse and a leg of one triangle are congruent to the corresponding parts of the other, then the two right triangles are congruent. (HL) Given Right ABC with right angle B and right DEF AC with right angle E, DF EF BC, C F Prove ABC DEF Proof To prove this theorem, we will construct a third triangle, GEF, that shares a common side with DEF and prove that each of the two given triangles is congruent to GEF and, thus, to each other. D A B E We first show that ABC is congruent to GEF: 14365C09.pgs 7/10/07 8:48 AM Page 363 Proving Right Triangles Congruent by Hypotenuse, Leg 363 (1) Since any line segment may be C F EG > AB to G so that extended any required length, extend DE FG (2) GEF and DEF form a linear pair, and DEF is a right angle. Therefore, GEF is a right angle. We are given that B is a right angle. All right angles are congruent, so B GEF.. Draw D A E B. G BC > EF (3) We are also given (4) Therefore, ABC GEF by SAS. We now show that DEF is also congruent to the constructed triangle, GEF:. (5) Since corresponding sides of congru- C F ent triangles are congruent, AC > GF AC > DF tive property of congruence.. Since we are given GF > DF, by the transi- A B D E G (6) If two sides of a triangle are congru- ent, the angles opposite these sides are congruent. In DFG, GF > DF, so D G. Also, DEF GE |
F since all right angles are congruent. (7) Therefore, DEF GEF by AAS. (8) Therefore, ABC DEF by the transitive property of congruence (steps 4 and 7). This theorem is called the hypotenuse-leg triangle congruence theorem, abbreviated HL. Therefore, from this point on, when the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of a second right triangle, we may say that the triangles are congruent. A corollary of this theorem is the converse of Corollary 9.12b. Corollary 9.14a If a point is equidistant from the sides of an angle, then it lies on the bisector of the angle. Given ABC, h PD'BA at D, h PF'BC at F, and PD PF Prove ABP CBP Strategy Use HL to prove PDB PFB. A D P The proof of this theorem is left to the stu- B F C dent. (See exercise 8.) 14365C09.pgs 7/10/07 8:48 AM Page 364 364 Parallel Lines Concurrence of Angle Bisectors of a Triangle In earlier chapters, we saw that the perpendicular bisectors of the sides of a triangle intersect in a point and that the altitudes of a triangle intersect in a point. Now we can prove that the angle bisectors of a triangle intersect in a point. Theorem 9.15 The angle bisectors of a triangle are concurrent. Given ABC with AL the bisector of B, and of C. the bisector of A, CN BM the bisector C Prove, BMAL, and CN intersect in a point, P. M L Proof Let P be the point at which P AL and BM intersect. If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. Therefore, P is equidisbecause it lies on the bisector of A, and P is equidistant from AB because it lies on the bisector of B. Therefore, P is tant from BC, equidistant from. If a point is equidistant from the sides of AC AB an angle, then it lies on the bisector of the angle. Since P is equidistant from AC bisectors of ABC intersect at a point, P., then it lies of the bisector of C. Therefore, the |
three angle and and AC AB, and and BC BC A N B The point where the angle bisectors of a triangle are concurrent is called the incenter. EXAMPLE 1 Given: ABC, AD AB. BC ⊥, BD AB DC, and C Prove: DAB BCD Proof We can show that ADB and CBD are right triangles and use HL to prove them congruent. A D B 14365C09.pgs 7/10/07 8:48 AM Page 365 Proving Right Triangles Congruent by Hypotenuse, Leg 365 Statements Reasons 1. 2. 3. ⊥ AB BD AB DC ⊥ BD DC 1. Given. 2. Given. 3. If a line is perpendicular to one of two parallel lines it is perpendicular to the other. 4. ABD and CDB are right 4. Perpendicular lines intersect to angles. 5. AD BC form right angles. 5. Given. BD 6. BD 7. ADB CBD 8. DAB BCD 6. Reflexive property of congruence. 7. HL (steps 5 and 6). 8. Corresponding parts of congruent triangles are congruent Exercises Writing About Mathematics 1. In two right triangles, the right angles are congruent. What other pairs of corresponding parts must be known to be congruent in order to prove these two right triangles congruent? 2. The incenter of ABC is P. If PD is the distance from P to AB and Q is any other point on AB, is PD greater than PQ, equal to PQ, or less than PQ? Justify your answer. Developing Skills 3. In ABC, mCAB 40 and mABC 60. The angle bisectors of ABC intersect at P. a. Find mBCA. b. Find the measure of each angle of APB. c. Find the measure of each angle of BPC. d. Find the measure of each angle of CPA. e. Does the bisector of CAB also bisect CPB? Explain your answer. M C P L A N B 14541C09.pgs 1/25/08 3:52 PM Page 366 366 Parallel Lines 4. Triangle ABC is an isosceles right triangle with the right angle at C. Let P be the incenter of ABC. a. Find the measure of each acute angle of ABC. b. Find the measure of each |
angle of APB. c. Find the measure of each angle of BPC. d. Find the measure of each angle of CPA. e. Does the bisector of ACB also bisect APB? Explain your answer. 5. Triangle ABC is an isosceles triangle with mC 140. Let P be the incenter of ABC. a. Find the measure of each acute angle of ABC. b. Find the measure of each angle of APB. c. Find the measure of each angle of BPC. d. Find the measure of each angle of CPA. e. Does the bisector of ACB also bisect APB? Explain your answer. 6. In RST, the angle bisectors intersect at P. If mRTS = 50, mTPR = 120, and mRPS 115, find the measures of TRS, RST, and SPT. 7. a. Draw a scalene triangle on a piece of paper or using geometry software. Label the trian- gle ABC. b. Using compass and straightedge or geometry software, construct the angle bisectors of the angles of the triangle. Let and be the bisector of A, be the bisector of C, such that L, M, and N are points on the triangle. be the bisector of B, BM CN AL c. Label the incenter P. d. In ABC, does AP = BP = CP? Explain why or why not. e. If the incenter is equidistant from the vertices of DEF, what kind of a triangle is DEF? Applying Skills 8. Prove Corollary 9.14a, “If a point is equidistant from the sides of an angle, then it lies on the bisector of the angle.” 9. Given DB'ABC and AD'DC, when is ABD congruent to DBC? Explain. D A B C 10. When we proved that the bisectors of the angles of a triangle intersect in a point, we began AL by stating that two of the angle bisectors, intersect, show that they are not parallel. (Hint: Show that a pair of interior angles on the same side of the transversal cannot be supplementary.), intersect at P. To prove that they. are cut by transversal AB and and BM BM AL 14365C09.pgs 7/10/07 8:48 AM Page 367 Interior |
and Exterior Angles of Polygons 367 11. Given: Quadrilateral ABCD, AB'BD, BD'DC, and AD > CB. Prove: A C and AD CB 12. In QRS, the bisector of QRS is perpendicular to QS at P. a. Prove that QRS is isosceles. b. Prove that P is the midpoint of QS. 13. Each of two lines from the midpoint of the base of an isosceles triangle is perpendicular to one of the legs of the triangle. Prove that these lines are congruent. 14. In quadrilateral ABCD, A and C are right angles and AB = CD. Prove that: a. AD = BC b. ABD CDB c. ADC is a right angle. 15. In quadrilateral ABCD, ABC and BCD are right angles, and AC = BD. Prove that AB = CD. 16. Given: Prove: h h, APS ABC h PD'ADE h APS bisects CAE. h with, and ADE, and PB PD. h PB'ABC, E D S P A B C 9-8 INTERIOR AND EXTERIOR ANGLES OF POLYGONS Polygons Recall that a polygon is a closed figure that is the union of line segments in a plane. Each vertex of a polygon is the endpoint of two line segments. We have proved many theorems about triangles and have used what we know about triangles to prove statements about the sides and angles of quadrilaterals, polygons with four sides. Other common polygons are: • A pentagon is a polygon that is the union of five line segments. • A hexagon is a polygon that is the union of six line segments. • An octagon is a polygon that is the union of eight line segments. • A decagon is a polygon that is the union of ten line segments. • In general, an n-gon is a polygon with n sides. 14365C09.pgs 7/10/07 8:48 AM Page 368 368 Parallel Lines A convex polygon is a polygon in which each of the interior angles measures less than 180 degrees. Polygon PQRST is a convex polygon and a pentagon. A concave polygon is a polygon in which at least one interior angle measures more than |
180 degrees. Polygon ABCD is a concave polygon and a quadrilateral. In the rest of this textbook, unless otherwise stated, all polygons are convex Interior Angles of a Polygon A pair of angles whose vertices are the endpoints of a common side are called consecutive angles. And the vertices of consecutive angles are called consecutive vertices or adjacent vertices. For example, in PQRST, P and Q are consecutive angles and P and Q are consecutive or adjacent vertices. Another pair of consecutive angles are T and P. Vertices R and T are nonadjacent vertices. A diagonal of a polygon is a line segment whose endpoints are two nonadjacent vertices. In hexagon ABCDEF, the vertices adjacent to B are A and C and the vertices nonadjacent to B are D, E, and F. Therefore, there are three diago. nals with endpoint B: BF, and, BD BE E F A B D C The polygons shown above have four, five, and six sides. In each polygon, all possible diagonals from a vertex are drawn. In the quadrilateral, two triangles are formed. In the pentagon, three triangles are formed, and in the hexagon, four triangles are formed. Note that in each polygon, the number of triangles formed is two less than the number of sides. • In a quadrilateral: the sum of the measures of the angles is 2(180) 360. • In a pentagon: the sum of the measures of the angles is 3(180) 540. • In a hexagon: the sum of the measures of the angles is 4(180) 720. 14365C09.pgs 7/10/07 8:48 AM Page 369 Interior and Exterior Angles of Polygons 369 In general, the number of triangles into which the diagonals from a vertex separate a polygon of n sides is two less than the number of sides, or n 2. The sum of the interior angles of the polygon is the sum of the interior angles of the triangles formed, or 180(n 2). We have just proved the following theorem: Theorem 9.16 The sum of the measures of the interior angles of a polygon of n sides is 180(n 2)°. Exterior Angles of a Polygon At any vertex of a polygon, an exterior angle forms a linear pair with the |
interior angle. The interior angle and the exterior angle are supplementary. Therefore, the sum of their measures is 180°. If a polygon has n sides, the sum of the interior and exterior angles of the polygon is 180n. Therefore, in a polygon with n sides: The measures of the exterior angles 180n the measures of the interior angles 180n 180(n 2) 180n 180n 360 360 We have just proved the following theorem: Theorem 9.17 The sum of the measures of the exterior angles of a polygon is 360°. DEFINITION A regular polygon is a polygon that is both equilateral and equiangular. If a triangle is equilateral, then it is equiangular. For polygons that have more than three sides, the polygon can be equiangular and not be equilateral, or can be equilateral and not be equiangular. Equilateral but not equiangular Equiangular but not equilateral Equiangular but not equilateral Equilateral but not equiangular 14365C09.pgs 7/10/07 8:48 AM Page 370 370 Parallel Lines EXAMPLE 1 The measure of an exterior angle of a regular polygon is 45 degrees. a. Find the number of sides of the polygon. b. Find the measure of each interior angle. c. Find the sum of the measures of the interior angles. Solution a. Let n be the number of sides of the polygon. Then the sum of the measures of the exterior angles is n times the measure of one exterior angle. 45n 5 360 n 5 360 45 n 5 8 Answer b. Each interior angle is the supplement of each exterior angle. Measure of each interior angle 180 45 135 Answer c. Use the sum of the measures of the interior angles, 180(n 2). 180(n 2 2) 5 180(8 2 2) 5 180(6) 5 1,080 Answer or Multiply the measure of each interior angle by the number of sides. 8(135) 1,080 Answer Answers a. 8 sides b. 135° c. 1,080° EXAMPLE 2 In quadrilateral ABCD, mA x, mB 2x 12, mC = x 22, and mD 3x. a. Find the measure of each interior angle of the quadrilateral. b. Find the measure of each exterior angle of the quadrilateral. Solution a. mA mB mC m |
D 18(n 2) x 2x 12 x 22 3x 180(4 2) 7x 10 360 7x 350 x 50 14365C09.pgs 7/10/07 8:48 AM Page 371 Interior and Exterior Angles of Polygons 371 mA x 50 mC x 22 50 22 72 mB 2x 12 2(50) 12 88 mD 3x 3(50) 150 b. Each exterior angle is the supplement of the interior angle with the same vertex. The measure of the exterior angle at A is 180 50 130. The measure of the exterior angle at B is 180 88 92. The measure of the exterior angle at C is 180 72 108. The measure of the exterior angle at D is 180 150 30. Answers a. 50°, 88°, 72°, 150° b. 130°, 92°, 108°, 30° Exercises Writing About Mathematics 1. Taylor said that each vertex of a polygon with n sides is the endpoint of (n 3) diagonals. Do you agree with Taylor? Justify your answer. n 2(n 2 3) 2. Ryan said that every polygon with n sides has Justify your answer. diagonals. Do you agree with Ryan? Developing Skills 3. Find the sum of the degree measures of the interior angles of a polygon that has: a. 3 sides b. 7 sides c. 9 sides d. 12 sides 4. Find the sum of the degree measures of the interior angles of: a. a hexagon b. an octagon c. a pentagon d. a quadrilateral 5. Find the sum of the measures of the exterior angles of a polygon that has: a. 4 sides b. 8 sides c. 10 sides d. 36 sides In 6–14, for each regular polygon with the given number of sides, find the degree measures of: a. one exterior angle b. one interior angle 6. 4 sides 9. 8 sides 12. 20 sides 7. 5 sides 10. 9 sides 13. 36 sides 8. 6 sides 11. 12 sides 14. 42 sides 14365C09.pgs 7/10/07 8:48 AM Page 372 372 Parallel Lines 15. Find the number of sides of a regular polygon each of whose exterior angles contains: a. 30° b. 45° c. 60° d. 120° 16. Find the number of sides of a regular polygon each of whose |
interior angles contains: a. 90° b. 120° c. 140° d. 160° 17. Find the number of sides a polygon if the sum of the degree measures of its interior angles is: a. 180 e. 1,440 b. 360 f. 2,700 c. 540 g. 1,800 d. 900 h. 3,600 Applying Skills 18. The measure of each interior angle of a regular polygon is three times the measure of each exterior angle. How many sides does the polygon have? 19. The measure of each interior angle of a regular polygon is 20 degrees more than three times the measure of each exterior angle. How many sides does the polygon have? 20. The sum of the measures of the interior angles of a concave polygon is also 180(n 2), where n is the number of sides. Is it possible for a concave quadrilateral to have two interior angles that are both more than 180°? Explain why or why not. 21. From vertex A of regular pentagon ABCDE, two diagonals are drawn, forming three triangles. a. Prove that two of the triangles formed by the diagonals are congruent. b. Prove that the congruent triangles are isosceles. c. Prove that the third triangle is isosceles. 22. From vertex L of regular hexagon LMNRST, three diagonals are drawn, forming four triangles. a. Prove that two of the triangles formed by the diagonals are congruent. b. Prove that the other two triangles formed by the diagonals are congruent. c. Find the measures of each of the angles in each of the four triangles. 23. The coordinates of the vertices of quadrilateral ABCD are A(2, 0), B(0, 2), C(2, 0), and D(0, 2). a. Prove that each angle of the quadrilateral is a right angle. b. Segments of the x-axis and the y-axis are diagonals of the quadrilateral. Prove that the four triangles into which the diagonals separate the quadrilateral are congruent. c. Prove that ABCD is a regular quadrilateral. Hands-On Activity In Section 9-7, we saw that the angle bisectors of a triangle are concurrent in a point called the incenter. In this |
activity, we will study the intersection of the angle bisectors of polygons. a. Draw various polygons that are not regular of different sizes and numbers of sides. Construct the angle bisector of each interior angle. Do the angle bisectors appear to intersect in a single point? b. Draw various regular polygons of different sizes and numbers of sides. Construct the angle bisector of each interior angle. Do the angle bisectors appear to intersect in a single point? c. Based on the results of part a and b, state a conjecture regarding the intersection of the angle bisector of polygons. 14365C09.pgs 7/10/07 8:48 AM Page 373 Chapter Summary 373 CHAPTER SUMMARY Definitions to Know • Parallel lines are coplanar lines that have no points in common, or have all points in common and, therefore, coincide. • A transversal is a line that intersects two other coplanar lines in two dif- ferent points. • The incenter is the point of intersection of the bisectors of the angles of a triangle. • A convex polygon is a polygon in which each of the interior angles mea- sures less than 180 degrees. • A concave polygon is a polygon in which at least one of the interior angles measures more than 180 degrees. • A regular polygon is a polygon that is both equilateral and equiangular. Postulates Theorems and Corollaries 9.1 9.2 9.1 9.2 9.3 9.4 9.5 9.6 9.7 Two distinct coplanar lines are either parallel or intersecting. Through a given point not on a given line, there exists one and only one line parallel to the given line. Two coplanar lines cut by a transversal are parallel if and only if the alternate interior angles formed are congruent. Two coplanar lines cut by a transversal are parallel if and only if corresponding angles are congruent. Two coplanar lines cut by a transversal are parallel if and only if interior angles on the same side of the transversal are supplementary. If two coplanar lines are each perpendicular to the same line, then they are parallel. If, in a plane, a line intersects one of two parallel lines, it intersects the other. If a transversal is perpendicular to one of two parallel lines, it is perpendicular |
to the other. If two of three lines in the same plane are each parallel to the third line, then they are parallel to each other. If two lines are vertical lines, then they are parallel. If two lines are horizontal lines, then they are parallel. 9.8 9.9 9.10 Two non-vertical lines in the coordinate plane are parallel if and only if they have the same slope. 9.11 The sum of the measures of the angles of a triangle is 180°. 9.11a If two angles of one triangle are congruent to two angles of another tri- angle, then the third angles are congruent. 9.11b The acute angles of a right triangle are complementary. 9.11c Each acute angle of an isosceles right triangle measures 45°. 9.11d Each angle of an equilateral triangle measures 60°. 9.11e The sum of the measures of the angles of a quadrilateral is 360°. 14365C09.pgs 7/10/07 8:48 AM Page 374 374 Parallel Lines 9.11f The measure of an exterior angle of a triangle is equal to the sum of the 9.12 measures of the nonadjacent interior angles. If two angles and the side opposite one of them in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent. (AAS) 9.12a Two right triangles are congruent if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of the other right triangle. 9.12b If a point lies on the bisector of an angle, then it is equidistant from the 9.13 sides of the angle. If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 9.13a If a triangle is equiangular, then it is equilateral. 9.14 If the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of the other, then the two right triangles are congruent. (HL) 9.14a If a point is equidistant from the sides of an angle, then it lies on the bisector of the angle. 9.15 The angle bisectors of a triangle are concurrent. 9.16 The sum of the measures of the interior angles of a polygon of |
n sides is 180(n 2)°. 9.17 The sum of the measures of the exterior angles of a polygon is 360°. VOCABULARY 9-1 Euclid’s parallel postulate • Playfair’s postulate • Coplanar • Parallel lines • Transversal • Interior angles • Exterior angles • Alternate interior angles • Alternate exterior angles • Interior angles on the same side of the transversal • Corresponding angles 9-3 Midsegment 9-5 AAS triangle congruence 9-7 Hypotenuse-leg triangle congruence theorem (HL) • Incenter 9-8 Pentagon • Hexagon • Octagon • Decagon • n-gon • Convex polygon • Concave polygon • Consecutive angles • Consecutive vertices • Adjacent vertices • Diagonal of a polygon • Regular polygon 14365C09.pgs 7/10/07 8:48 AM Page 375 REVIEW EXERCISES In 1–5, g AB g CD respectively. and these lines are cut by transversal 1. If mAEF 5x and mDFE 75, find x. 2. If mCFE 3y 20 and mAEG 4y 10, find y. 3. If mBEF 5x and mCFE 7x 48, find x. 4. If mDFE y and mBEF 3y 40, find mDFE. A C Review Exercises 375 at points E and F, g GH G E B F D H 5. If mAEF 4x and mEFD 3x 18, find: a. the value of x c. mEFD b. mAEF d. mBEF e. mCFH 6. The degree measure of the vertex angle of an isosceles triangle is 120. Find the measure of a base angle of the triangle. 7. In ABC, A C. If AB 8x 4 and CB 3x 34, find x. 8. In an isosceles triangle, if the measure of the vertex angle is 3 times the measure of a base angle, find the degree measure of a base angle. 9. In a triangle, the degree measures of the three angles are represented by x, x 42, and x 6. Find the angle measures. 10. In PQR, if mP 35 and mQ 85, what is the degree measure |
of an exterior angle of the triangle at vertex R? 11. An exterior angle at the base of an isosceles triangle measures 130°. Find the measure of the vertex angle. 12. In ABC, if 13. In DEF, if 14. In PQR, AB DE AC DF and mA 70, find mB. and mE 13, find mD. PQ is extended through Q to point T, forming exterior RQT. If mRQT 70 and mR 10, find mP. 15. In ABC, AC BC. The degree measure of an exterior angle at vertex C is represented by 5x 10. If mA 30, find x. 16. The degree measures of the angles of a triangle are represented by x 10, 2x 20, and 3x 10. Find the measure of each angle of the triangle. 17. If the degree measures of the angles of a triangle are represented by x, y, and x y, what is the measure of the largest angle of the triangle? 14365C09.pgs 7/10/07 8:48 AM Page 376 376 Parallel Lines 18. If parallel lines are cut by a transversal so that the degree measures of two corresponding angles are represented by 2x 50 and 3x 20, what is the value of x? 19. The measure of one exterior angle of a regular polygon is 30°. How many sides does the regular polygon have? 20. What is the sum of the degree measures of the interior angles of a polygon with nine sides? 21. Given: Right triangle ABC with C the right angle. Prove: AB AC 22. Given: Prove: and AEB AC BD CED bisect each other at E. 23. P is not on ABCD APB DPC, prove that and PA PB PC PA > PD, and.,, PD are drawn. If PB PC and 24. P is not on ABCD and AB > DC,, and PA PB PA > PD, prove that, and. PC PD are drawn. If PBC PCB 25. Herbie wanted to draw pentagon ABCDE with mA = mB 120 and mC = mD = 150. Is such a pentagon possible? Explain your answer. Exploration The geometry that you have been studying is called plane Euclidean geometry. Investigate a non-Euclidean geometry. How do the postulates of a nonEuclidean geometry differ |
from the postulates of Euclid? How can the postulates from this chapter be rewritten to fit the non-Euclidean geometry you investigated? What theorems from this chapter are not valid in the nonEuclidean geometry that you investigated? One possible non-Euclidean geometry is the geometry of the sphere suggested in the Chapter 1 Exploration. CUMULATIVE REVIEW Chapters 1–9 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed., which of the following may be false? 1. If M is the midpoint of AB (1) M is between A and B. (2) AM = MB (3) A, B, and M are collinear. g MN (4), a line that intersects AB at M, is the perpendicular bisector of.AB 14365C09.pgs 7/10/07 8:48 AM Page 377 Cumulative Review 377 2. The statement “If two angles form a linear pair, then they are supplemen- tary” is true. Which of the following statements must also be true? (1) If two angles do not form a linear pair, then they are not supplementary. (2) If two angles are not supplementary, then they do not form a linear pair. (3) If two angles are supplementary, then they form a linear pair. (4) Two angles form a linear pair if and only if they are supplementary. 3. Which of the following is a statement of the reflexive property of equality for all real numbers a, b, and c? (1) a = a (2) If a = b, then b = a. (3) If a = b and b = c, then a = c. (4) If a = b, then ac = bc. 4. Two angles are complementary. If the measure of the larger angle is 10 degrees less than three times the measure of the smaller, what is the measure of the larger angle? (1) 20° (2) 25° (3) 65° (4) 70° 5. Under the transformation + R908, the image of (2, 5) is (1) (5, 2) (4) (2, 5) 6. An equation of the line through (0, 1) and perpendicular to the line (3) (5, 2) rx-axis (2 |
) (5, 2) x 3y 4 is (1) 3x y 1 (2) x 3y 1 (3) 3x y 1 (4) x 3y 1 7. The coordinates of the midpoint of the line segment whose endpoints are (3, 4) and (5, 6) are (1) (1, 1) (2) (4, 5) (3) (4, 5) (4) (4, 5) 8. If a, b, c, and d are real numbers and a b and c d, which of the follow- ing must be true? (1) a c b d (2) a c b d (3) ac bc c. b a (4) d 9. The measure of each base angle of an isosceles triangle is 5 more than twice the measure of the vertex angle. The measure of the vertex angle is (1) 34° (4) 136.25° (3) 43.75° (2) 73° 10. Which of the following properties is not preserved under a line reflection? (1) distance (2) orientation (3) angle measure (4) midpoint 14365C09.pgs 7/10/07 8:48 AM Page 378 378 Parallel Lines 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. g AD. If mCAB 65, and B is a point that is not on mCBD = 20, and mBCD 135, which is the longest side of ABC?, prove that ABP is 12. If P is a point on the perpendicular bisector of 11. C is a point on AD AB isosceles. 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. ABCD is an equilateral quadrilateral. Prove that the diagonal, AC, bisects DAB and DCB. g AEB g CED g CB and g AD 14. intersect at E and Prove that mDEB mEBC |
mEDA.. A D E C B 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 a triangle are in the ratio 3 : 4 : 8. Find the measure of the smallest exterior angle. 16. Write an equation of the perpendicular bisector of are A(1, 2) and B(7, 6). of the endpoints of AB AB if the coordinates 14365C10.pgs 7/10/07 8:50 AM Page 379 QUADRILATERALS Euclid’s fifth postulate was often considered to be a “flaw” in his development of geometry. Girolamo Saccheri (1667–1733) was convinced that by the application of rigorous logical reasoning, this postulate could be proved. He proceeded to develop a geometry based on an isosceles quadrilateral with two base angles that are right angles.This isosceles quadrilateral had been proposed by the Persian mathematician Nasir al-Din al-Tusi (1201–1274). Using this quadrilateral, Saccheri attempted to prove Euclid’s fifth postulate by reasoning to a contradiction. After his death, his work was published under the title Euclid Freed of Every Flaw. Saccheri did not, as he set out to do, prove the parallel postulate, but his work laid the foundations for new geometries. János Bolyai (1802–1860) and Nicolai Lobachevsky (1793–1856) developed a geometry that allowed two lines parallel to the given line, through a point not on a given line. Georg Riemann (1826–1866) developed a geometry in which there is no line parallel to a given line through a point not on the given line. CHAPTER 10 CHAPTER TABLE OF CONTENTS 10-1 The General Quadrilateral 10-2 The Parallelogram 10-3 Proving That a Quadrilateral Is a Parallelogram 10-4 The Rectangle 10-5 The Rhombus 10-6 The Square 10-7 The Trapezoid 10-8 Areas of Polygons Chapter Summary Vocabulary Review Ex |
ercises Cumulative Review 379 14365C10.pgs 7/10/07 8:50 AM Page 380 380 Quadrilaterals 10-1 THE GENERAL QUADRILATERAL Patchwork is an authentic American craft, developed by our frugal ancestors in a time when nothing was wasted and useful parts of discarded clothing were stitched into warm and decorative quilts. Quilt patterns, many of which acquired names as they were handed down from one generation to the next, were the product of creative and industrious people. In the Lone Star pattern, quadrilaterals are arranged to form larger quadrilaterals that form a star. The creators of this pattern were perhaps more aware of the pleasing effect of the design than of the mathematical relationships that were the basis of the pattern. A quadrilateral is a polygon with four sides. In this chapter we will study the various special quadrilaterals and the properties of each. Let us first name the general parts and state properties of any quadrilateral, using PQRS as an example. • Consecutive vertices or adjacent vertices are vertices that are endpoints of the same side such as P and Q, Q and R, R and S, S and P. Q P R S • Consecutive sides or adjacent sides are sides that have a common endand point, such as and and and, SP QR QR PQ PQ RS RS SP,., • Opposite sides of a quadrilateral are sides that do not have a common endpoint, such as PQ and RS, SP and QR. • Consecutive angles of a quadrilateral are angles whose vertices are consec- utive, such as P and Q, Q and R, R and S, S and P. • Opposite angles of a quadrilateral are angles whose vertices are not con- secutive, such as P and R, Q and S. • A diagonal of a quadrilateral is a line segment whose endpoints are two nonadjacent vertices of the quadrilateral, such as PR and QS. • The sum of the measures of the angles of a quadrilateral is 360 degrees. Therefore, mP mQ mR mS 360. 10-2 THE PARALLELOGRAM DEFINITION A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. 14365C10.pgs 7/10/07 8:50 AM Page 38 |
1 D C Quadrilateral ABCD is a parallelogram because The symbol for parallelogram ABCD is ~ABCD. The Parallelogram 381 AB CD and BC DA. A B that are parallel in the figure. Note the use of arrowheads, pointing in the same direction, to show sides Theorem 10.1 A diagonal divides a parallelogram into two congruent triangles. Given Parallelogram ABCD with diagonal AC D C Prove ABC CDA A B Proof Since opposite sides of a parallelogram are parallel, alternate interior angles can be proved congruent using the diagonal as the transversal. Statements Reasons 1. ABCD is a parallelogram. 1. Given. 2. AB CD and BC DA 3. BAC DCA and BCA DAC AC > AC 4. 5. ABC CDA 2. A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. 3. If two parallel lines are cut by a transversal, alternate interior angles are congruent. 4. Reflexive property of congruence. 5. ASA. We have proved that the diagonal AC two congruent triangles. An identical proof could be used to show that divides the parallelogram into two congruent triangles, ABD CDB. divides parallelogram ABCD into BD The following corollaries result from this theorem. Corollary 10.1a Opposite sides of a parallelogram are congruent. Corollary 10.1b Opposite angles of a parallelogram are congruent. The proofs of these corollaries are left to the student. (See exercises 14 and 15.) 14365C10.pgs 7/10/07 8:50 AM Page 382 382 Quadrilaterals We can think of each side of a parallelogram as a segment of a transversal that intersects a pair of parallel lines. This enables us to prove the following theorem. Theorem 10.2 Two consecutive angles of a parallelogram are supplementary. Proof ~ABCD, opposite sides are parallel. If In two parallel lines are cut by a transversal, then two interior angles on the same side of the transversal are supplementary. Therefore, A is supplementary to B, B is supplementary to C, C is supplementary to D, and D is supplementary to A. D C A B Theorem 10.3 The diagonals of a parallelog |
ram bisect each other. Given ~ABCD with diagonals AC and BD intersecting at E. D C Prove AC and BD bisect each other. E A B Proof Statements Reasons 1. AB CD D C 2. BAE DCE and ABE CDE B 3. AB > CD C 4. ABE CDE E E D A A 1. Opposite sides of a parallelogram are parallel. 2. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 3. Opposite sides of a parallelogram are congruent. 4. ASA. 5. AE > CE and BE > DE 5. Corresponding part of congruent triangles are congruent. B 6. E is the midpoint of AC and 6. The midpoint of a line segment of BD. divides the segment into two congruent segments. 7. AC and BD bisect each other. 7. The bisector of a line segment intersects the segment at its midpoint. 14365C10.pgs 7/10/07 8:50 AM Page 383 DEFINITION The distance between two parallel lines is the length of the perpendicular from any point on one line to the other line. The Parallelogram 383 Properties of a Parallelogram 1. Opposite sides are parallel. 2. A diagonal divides a parallelogram into two congruent triangles. 3. Opposite sides are congruent. 4. Opposite angles are congruent. 5. Consecutive angles are supplementary. 6. The diagonals bisect each other. EXAMPLE 1 In ~ABCD, mB exceeds mA by 46 degrees. Find mB. Solution Let x mA. Then x 46 mB. Two consecutive angles of a parallelogram are supplementary. Therefore, mA mB 180 x x 46 180 2x 46 180 2x 134 x 67 mB x 46 67 46 113 Answer mB 113 Exercises Writing About Mathematics 1. Theorem 10.2 states that two consecutive angles of a parallelogram are supplementary. If two opposite angles of a quadrilateral are supplementary, is the quadrilateral a parallelogram? Justify your answer. 2. A diagonal divides a parallelogram into two congruent triangles. Do two diagonals divide a parallelogram into four congruent triangles? Justify your answer. 14365C10.pgs 7/10 |
/07 8:50 AM Page 384 384 Quadrilaterals Developing Skills 3. Find the degree measures of the other three angles of a parallelogram if one angle measures: a. 70 b. 65 c. 90 d. 130 e. 155 f. 168 In 4–11, ABCD is a parallelogram. 4. The degree measure of A is represented by 2x 20 and the degree measure of B by 2x. Find the value of x, of mA, and of mB. 5. The degree measure of A is represented by 2x 10 and the degree measure of B by 3x. Find the value of x, of mA, and of mB. D C A B 6. The measure of A is 30 degrees less than twice the measure of B. Find the measure of each angle of the parallelogram. 7. The measure of A is represented by x 44 and the measure of C by 3x. Find the mea- sure of each angle of the parallelogram. 8. The measure of B is represented by 7x and mD by 5x 30. Find the measure of each angle of the parallelogram. 9. The measure of C is one-half the measure of B. Find the measure of each angle of the parallelogram. 10. If AB 4x 7 and CD 3x 12, find AB and CD. 11. If AB 4x y, BC y 4, CD 3x 6, and DA 2x y, find the lengths of the sides of the parallelogram. 12. The diagonals of 13. The diagonals of ~ABCD ~ABCD intersect at E. If AE 5x 3 and EC 15 x, find AC. intersect at E. If DE 4y 1 and EB 5y 1, find DB. Applying Skills 14. Prove Corollary 10.1a, “Opposite sides of a parallelogram are congruent.” 15. Prove Corollary 10.1b, “Opposite angles of a parallelogram are congruent.” 16. Given: Parallelogram EBFD and parallelogram ABCD with D C F EAB and DCF Prove: EAD FCB AE B 17. Petrina said that the floor of her bedroom is in the shape of a parallelogram and that at least one of the angles is a right angle. Show that the floor of |
Petrina’s bedroom has four right angles. 18. The deck that Jeremiah is building is in the shape of a quadrilateral, ABCD. The measure of the angle at A is not equal to the measure of the angle at C. Prove that the deck is not in the shape of a parallelogram. 14365C10.pgs 7/10/07 8:50 AM Page 385 Proving That a Quadrilateral Is a Parallelogram 385 19. Quadrilaterals ABCD and PQRS are parallelograms with. Prove that ABCD PQRS or draw a counterexample to show that they may not be congruent. AB > PQ BC > QR and 20. Quadrilaterals ABCD and PQRS are parallelograms with AB > PQ, BC > QR, and B Q. Prove that ABCD PQRS or draw a counterexample to show that they may not be congruent. 10-3 PROVING THAT A QUADRILATERAL IS A PARALLELOGRAM If we wish to prove that a certain quadrilateral is a parallelogram, we can do so by proving its opposite sides are parallel, thus satisfying the definition of a parallelogram. Now we want to determine other ways of proving that a quadrilateral is a parallelogram. Theorem 10.4 If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with AB > CD, AD > BC D C Prove ABCD is a parallelogram. Proof AC is a diagonal. Triangles ABC and CDA In ABCD, are congruent by SSS. Corresponding parts of congruent triangles are congruent, so BAC DCA and DAC ACB. is a. Alternate interior angles BAC and DCA transversal that cuts DC is also a transversal that cuts are congruent, so AC Alternate interior angles DAC and ACB are congruent so Therefore, ABCD is a parallelogram. and AD AD BC AB AB DC and. BC. AC A B. Theorem 10.5 If one pair of opposite sides of a quadrilateral is both congruent and parallel, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with AB CD and AB > CD D C Prove ABC |
D is a parallelogram. Proof Since AB is parallel to CD, BAC and DCA are a A B pair of congruent alternate interior angles. Therefore, by SAS, DCA BAC. Corresponding parts of congruent triangles are congruent, so DAC ACB. and congruent alternate interior angles DAC and ACB. Therefore, and ABCD is a parallelogram. is a transversal that cuts BC AD BC AD AC,, forming 14365C10.pgs 7/10/07 8:50 AM Page 386 386 Quadrilaterals Theorem 10.6 If both pairs of opposite angles of a quadrilateral are congruent, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with A C and B D D C Prove ABCD is a parallelogram. Proof The sum of the measures of the angles of a quadrilateral B is 360 degrees. Therefore, mA mB mC mD 360. It is given that A C and B D. Congruent angles have equal measures so mA mC and mB mD. A By substitution, mA mD mA mD 360. Then, 2mA 2mD 360 or mA mD 180. Similarly, mA mB 180. If the sum of the measures of two angles is 180, the angles are supplementary. Therefore, A and D are supplementary and A and B are supplementary. Two coplanar lines are parallel if a pair of interior angles on the same side of the transversal are supplementary. Therefore, Quadrilateral ABCD is a parallelogram because it has two pairs of parallel sides. and. AD BC AB DC Theorem 10.7 If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with AE > EC, BE > ED Prove ABCD is a parallelogram. AC and BD intersecting at E, D C E A B Strategy Prove that ABE CDE to show that one pair of opposite sides is congruent and parallel. The proof of Theorem 10.7 is left to the student. (See exercise 15.) SUMMARY To prove that a quadrilateral is a parallelogram, prove that any one of the following statements is true: 1. Both pairs of opposite sides are parallel. 2. Both pairs of opposite sides are con |
gruent. 3. One pair of opposite sides is both congruent and parallel. 4. Both pairs of opposite angles are congruent. 5. The diagonals bisect each other. 14365C10.pgs 7/10/07 8:50 AM Page 387 EXAMPLE 1 Proving That a Quadrilateral Is a Parallelogram 387 Given: ABCD is a parallelogram. D F C E is on AB, F is on DC, and EB > DF. Prove: DE FB Proof We will prove that EBFD is a parallelogram. A E B Statements Reasons 1. ABCD is a parallelogram. 1. Given. 2. AB DC 3. EB DF 4. EB > DF 5. EBFD is a parallelogram. 6. DE FB 2. Opposite sides of a parallelogram are parallel. 3. Segments of parallel lines are parallel. 4. Given. 5. If one pair of opposite sides of a quadrilateral is both congruent and parallel, the quadrilateral is a parallelogram. 6. Opposite sides of a parallelogram are parallel. Exercises Writing About Mathematics 1. What statement and reason can be added to the proof in Example 1 to prove that DE > FB? 2. What statement and reason can be added to the proof in Example 1 to prove that DEB BFD? Developing Skills In 3–7, in each case, the given is marked on the figure. Tell why each quadrilateral ABCD is a parallelogram. 3. D C 4. D C 5. D C 6. D C 7 14365C10.pgs 7/10/07 8:50 AM Page 388 388 Quadrilaterals 8. ABCD is a quadrilateral with AB CD and A C. Prove that ABCD is a parallelo- gram. 9. PQRS is a quadrilateral with P R and P the supplement of Q. Prove that PQRS is a parallelogram. 10. DEFG is a quadrilateral with DF drawn so that FDE DFG and GDF EFD. Prove that DEFG is a parallelogram. 11. ABCD is a parallelogram. E is the midpoint of AB and F is the midpoint of CD. Prove that AEFD is a parallelogram. 12. EFGH is a parallelogram and J is a |
point on h EF such that F is the midpoint of EJ. Prove that FJGH is a parallelogram. CD 13. ABCD is a parallelogram. The midpoint of is R, and the midpoint of of a. Prove that APS CRQ and that BQP DSR. b. Prove that PQRS is a parallelogram. is S. DA AB is P, the midpoint of BC is Q, the midpoint 14. A quadrilateral has three right angles. Is the quadrilateral a parallelogram? Justify your answer. Applying Skills 15. Prove Theorem 10.7, “If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.” 16. Prove that a parallelogram can be drawn by joining the endpoints of two line segments that bisect each other. 17. The vertices of quadrilateral ABCD are A(2, 1), B(4, 2), C(8, 2), and D(2, 5). Is ABCD a parallelogram? Justify your answer. 18. Farmer Brown’s pasture is in the shape of a quadrilateral, PQRS. The pasture is crossed by two diagonal paths, QS paths do not bisect each other. and PR. The quadrilateral is not a parallelogram. Show that the 19. Toni cut two congruent scalene triangles out of cardboard. She labeled one triangle ABC and the other ABC so that ABC ABC. She placed the two triangles next to each other so that A coincided with B and B coincided with A. Was the resulting quadrilateral ACBC a parallelogram? Prove your answer. 20. Quadrilateral ABCD is a parallelogram with M the midpoint of AB and N the midpoint of CD. a. Prove that AMND and MBCN are parallelograms. b. Prove that AMND and MBCN are congruent. 14365C10.pgs 7/10/07 8:50 AM Page 389 10-4 THE RECTANGLE The Rectangle 389 DEFINITION A rectangle is a parallelogram containing one right angle. If one angle, A, of a parallelogram ABCD is a right angle, then ~ABCD is a rectangle. Any side of a rectangle may be called |
the base of the rectangle. Thus, if side, is called the is taken as the base, then either consecutive side, AD BC or AB altitude of the rectangle. Since a rectangle is a special kind of parallelogram, a rectangle has all the properties of a parallelogram. In addition, we can prove two special properties for the rectangle. Theorem 10.8 All angles of a rectangle are right angles. Given Rectangle ABCD with A a right angle. Prove B, C, and D are right angles. D A C B Proof By definition, rectangle ABCD is a parallelogram. Opposite angles of a parallelogram are congruent, so A C. Angle A is a right angle so C is a right angle. Consecutive angles of a parallelogram are supplementary. Therefore, since A and B are supplementary and A is right angle, B is also a right angle. Similarly, since C and D are supplementary and C is a right angle, D is also a right angle. Theorem 10.9 The diagonals of a rectangle are congruent. Given ABCD is a rectangle. Prove AC > BD Strategy Prove DAB CBA. D A C B The proof of Theorem 10.9 is left to the student. (See exercise 12.) 14365C10.pgs 7/10/07 8:50 AM Page 390 390 Quadrilaterals Properties of a Rectangle 1. A rectangle has all the properties of a parallelogram. 2. A rectangle has four right angles and is therefore equiangular. 3. The diagonals of a rectangle are congruent. Proving That a Quadrilateral Is a Rectangle We prove that a quadrilateral is a rectangle by showing that it has the special properties of a rectangle. For example: Theorem 10.10 If a quadrilateral is equiangular, then it is a rectangle. Given Quadrilateral ABCD with A B C D. D C Prove ABCD is a rectangle. Proof Quadrilateral ABCD is a parallelogram because the opposite angles are congruent. The sum of the measures of the angles of a quadrilateral is 360 degrees. Thus, the measure of each of the four congruent angles is 90 degrees. Therefore, ABCD is a rectangle because it is a parallelogram with a right angle. A B Theorem 10.11 If the diagonals of a parallelogram are |
congruent, the parallelogram is a rectangle. Given Parallelogram ABCD with AC > BD Prove ABCD is a rectangle. Strategy Prove that DAB CBA. Therefore, DAB and CBA are both congruent and supplementary. D A C B The proof of Theorem 10.11 is left to the student. (See exercise 13.) SUMMARY To prove that a quadrilateral is a rectangle, prove that any one of the following statements is true: 1. The quadrilateral is a parallelogram with one right angle. 2. The quadrilateral is equiangular. 3. The quadrilateral is a parallelogram whose diagonals are congruent. 14365C10.pgs 7/10/07 8:50 AM Page 391 EXAMPLE 1 Given: ABCD is a parallelogram with mA mD. Prove: ABCD is a rectangle. Proof We will prove that ABCD has a right angle. The Rectangle 391 D A C B Statements Reasons 1. ABCD is a parallelogram. 1. Given. 2. AB CD 3. A and D are supplementary. 4. mA mD 180 5. mA mD 6. mA mA 180 or 2mA 180 7. mA 90 2. A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. 3. Two consecutive angles of a parallelogram are supplementary. 4. Supplementary angles are two angles the sum of whose measures is 180. 5. Given. 6. A quantity may be substituted for its equal. 7. Division postulate. 8. ABCD is a rectangle. 8. A rectangle is a parallelogram with a right angle. EXAMPLE 2 The lengths of diagonals of a rectangle are represented by 7x centimeters and 5x 12 centimeters. Find the length of each diagonal. Solution The diagonals of a rectangle are congruent and therefore equal in length. 7x 5x 12 2x 12 x 6 7x 5 7(6) 5x 1 12 5 5(6) 1 12 5 42 5 30 1 12 5 42 Answer The length of each diagonal is 42 centimeters. 14365C10.pgs 7/10/07 8:50 AM Page 392 392 Quadrilaterals Exercises Writing About Mathematics 1. Pauli said that if one angle of a parallelogram is not a right angle |
, then the parallelogram is not a rectangle. Do you agree with Pauli? Explain why or why not. 2. Cindy said that if two congruent line segments intersect at their midpoints, then the quadrilateral formed by joining the endpoints of the line segments in order is a rectangle. Do you agree with Cindy? Explain why or why not. Developing Skills In 3–10, the diagonals of rectangle ABCD intersect at E. 3. Prove that ABE is isosceles. 4. AC 4x 6 and BD 5x 2. Find AC, BD, AE, and BE. 5. AE y 12 and DE 3y 8. Find AE, DE, AC, and BD. 6. BE 3a 1 and ED 6a 11. Find BE, ED, BD, and AC. 7. AE x 5 and BD 3x 2. Find AE, BD, AC, and BE. 8. mCAB 35. Find mCAD, mACB, mAEB, and AED. 9. mAEB 3x and mDEC x 80. Find mAEB, mDEC, mCAB, and mCAD. 10. mAED y 10 and mAEB 4y 30. Find mAED, mAEB, mBAC, and D A E mCAD. C B Applying Skills 11. Write a coordinate proof of Theorem 10.8, “All angles of a rectangle are right angles.” Let the vertices of the rectangle be A(0, 0), B(b, 0), C(b, c), and D(0, c). 12. Prove Theorem 10.9, “The diagonals of a rectangle are congruent.” 13. Prove Theorem 10.11, “If the diagonals of a parallelogram are congruent, the parallelogram is a rectangle.” 14. If PQRS is a rectangle and M is the midpoint of 15. The coordinates of the vertices of ABCD are A(2, 0), B(2, 2), C(5, 4), and D(1, 6)., prove that PM > QM RS. a. Prove that ABCD is a rectangle. b. What are the coordinates of the point of intersection of the diagonals? c. The vertices of ABCD are A( |
0, 2), B(2, 2), C(4, 5), and D(6, 1). Under what specific transformation is ABCD the image of ABCD? d. Prove that ABCD is a rectangle. 14365C10.pgs 7/10/07 8:50 AM Page 393 The Rhombus 393 16. The coordinates of the vertices of PQRS are P(2, 1), Q(1, 3), R(5, 1), and S(2, 5). a. Prove that PQRS is a parallelogram. b. Prove that PQRS is not a rectangle. c. PQRS is the image of PQRS under T–3,–3 PQRS? d. Prove that PQRS is congruent to PQRS. + ry = x. What are the coordinates of 17. Angle A in quadrilateral ABCD is a right angle and quadrilateral ABCD is not a rectangle. Prove that ABCD is not a parallelogram. 18. Tracy wants to build a rectangular pen for her dog. She has a tape measure, which enables her to make accurate measurements of distance, but has no way of measuring angles. She places two stakes in the ground to represent opposite corners of the pen. How can she find two points at which to place stakes for the other two corners? 19. Archie has a piece of cardboard from which he wants to cut a rectangle with a diagonal that measures 12 inches. On one edge of the cardboard Archie marks two points that are less than 12 inches apart to be the endpoints of one side of the rectangle. Explain how Archie can use two pieces of string that are each 12 inches long to find the other two vertices of the rectangle. 10-5 THE RHOMBUS DEFINITION A rhombus is a parallelogram that has two congruent consecutive sides. If the consecutive sides AB > AD ), then (that is, if and AB ~ABCD is a rhombus. AD of parallelogram ABCD are congruent Since a rhombus is a special kind of parallelogram, a rhombus has all the properties of a parallelogram. In addition, we can prove three special properties for the rhombus. Theorem 10.12 All sides of a rhombus are congruent. Given ABCD is a rhombus with AB > DA. D |
C Prove AB > BC > CD > DA Proof By definition, rhombus ABCD is a parallelogram. It is AB > DA given that and are congruent, so transitive property of congruence, AB > CD. Opposite sides of a parallelogram BC > DA. Using the AB > BC > CD > DA. A B 14365C10.pgs 7/10/07 8:50 AM Page 394 394 Quadrilaterals Theorem 10.13 The diagonals of a rhombus are perpendicular to each other. Given Rhombus ABCD Prove AC'BD D C Proof By Theorem 10.12, all sides of a rhombus are congruent. AC. Point B is equidistant from the endpoints Segments that are congruent are equal. Consider the diagonal A and C since BA BC. Point D is also equidistant from the endpoints A and C since DA DC. If two points are each equidistant from the endpoints of a line segment, the points determine the perpendicular bisector of the line segment. Therefore, AC'BD A B. Theorem 10.14 The diagonals of a rhombus bisect its angles. Given Rhombus ABCD Prove bisects DAB and DCB and AC and CBA. DB bisects CDA Strategy Show that the diagonals separate the rhombus into four A congruent triangles. D C E B The proof of this theorem is left to the student. (See exercise 16.) Properties of a Rhombus 1. A rhombus has all the properties of a parallelogram. 2. A rhombus is equilateral. 3. The diagonals of a rhombus are perpendicular to each other. 4. The diagonals of a rhombus bisect its angles. Methods of Proving That a Quadrilateral Is a Rhombus We prove that a quadrilateral is a rhombus by showing that it has the special properties of a rhombus. 14365C10.pgs 7/10/07 8:50 AM Page 395 Theorem 10.15 If a quadrilateral is equilateral, then it is a rhombus. The Rhombus 395 Given Quadrilateral ABCD with AB > BC > CD > DA D C Prove ABCD is a rhombus. Proof AB > BC > CD > DA |
It is given that in ABCD,. Since both pairs of opposite sides are congruent, ABCD is a parallelogram. Two consecutive sides of parallelogram ABCD are congruent, so by definition, ABCD is a rhombus. A B Theorem 10.16 If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. Given Parallelogram ABCD with AC'BD D C Prove ABCD is a rhombus. Strategy The diagonals divide parallelogram ABCD into four triangles. Prove that two of the triangles that share a common side are congruent. Then use the fact that corresponding parts of congruent triangles are congruent to show that parallelogram ABCD has two congruent consecutive sides. E A B The proof of this theorem is left to the student. (See exercise 17.) SUMMARY To prove that a quadrilateral is a rhombus, prove that any one of the following statements is true: 1. The quadrilateral is a parallelogram with two congruent consecutive sides. 2. The quadrilateral is equilateral. 3. The quadrilateral is a parallelogram whose diagonals are perpendicular to each other. EXAMPLE 1 PQRS is a rhombus and mPQR 60. Prove that divides the rhombus into two equilateral triangles. PR P Q 60° S R 14365C10.pgs 7/10/07 8:50 AM Page 396 396 Quadrilaterals Proof Since all sides of a rhombus are congruent, we know that PQ > RQ. Thus, PQR is isosceles and its base angles are equal in measure. Let mPRQ mRPQ x. x x 60 180 2x 60 180 2x 120 x 60 Therefore, mPRQ 60, mRPQ 60, and mPQR 60. Triangle PQR is equilateral since an equiangular triangle is equilateral. Since opposite angles of a rhombus have equal measures, mRSP 60. By similar reasoning, RSP is equilateral. P Q 60° S R EXAMPLE 2 Given: ABCD is a parallelogram. D C AB 2x 1, DC 3x 11, AD x 13. Prove: ABCD is a rhombus. Proof (1) |
Since ABCD is a parallelogram, opposite sides are equal in length: (2) Substitute x 12 to find the : AD lengths of sides and AB A B DC AB 3x 11 2x 1 x 12 AB 2x 1 2(12) 1 25 AD x 13 12 13 25 (3) Since ABCD is a parallelogram with two congruent consecutive sides, ABCD is a rhombus. Exercises Writing About Mathematics 1. Rochelle said that the diagonals of a rhombus separate the rhombus into four congruent right triangles. Do you agree with Rochelle? Explain why or why not. 14365C10.pgs 7/10/07 8:50 AM Page 397 The Rhombus 397 2. Concepta said that if the lengths of the diagonals of a rhombus are represented by d1 and. Do you agree with Concepta? d2, then a formula for the area of a rhombus is A Explain why or why not. 1 2d1d2 Developing Skills In 3–10, the diagonals of rhombus ABCD intersect at E. 3. Name four congruent line segments. 4. Name two pairs of congruent line segments. D C 5. Name a pair of perpendicular line segments. E 6. Name four right angles. 7. Under a rotation of 90o about E, does A map to B? does B map to C? Justify your answer. A B 8. Does rhombus ABCD have rotational symmetry under a rotation of 90o about E? 9. Under a reflection in E, name the image of A, of B, of C, of D, and of E. 10. Does rhombus ABCD have point symmetry under a reflection in E? 11. In rhombus PQRS, mP is 120. a. Prove that the diagonal b. If PR 24 cm, what is the length of each side of the rhombus? separates the rhombus into two equilateral triangles. PR In 12–15, tell whether each conclusion follows from the given premises. If not, draw a counterexample. 12. In a parallelogram, opposite sides are congruent. A rhombus is a parallelogram. Conclusion: In a rhombus, opposite sides are congruent. 13. In a rhombus, diagonals are |
perpendicular to each other. A rhombus is a parallelogram. Conclusion: In a parallelogram, diagonals are perpendicular to each other. 14. The diagonals of a rhombus bisect the angles of the rhombus. A rhombus is a parallelogram. Conclusion: The diagonals of a parallelogram bisect its angles. 15. Consecutive angles of a parallelogram are supplementary. A rhombus is a parallelogram. Conclusion: Consecutive angles of a rhombus are supplementary. Applying Skills 16. Prove Theorem 10.14, “The diagonals of a rhombus bisect its angles.” 17. Prove Theorem 10.16, “If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.” 14365C10.pgs 7/10/07 8:50 AM Page 398 398 Quadrilaterals 18. The vertices of quadrilateral ABCD are A(1, 1), B(4, 0), C(5, 5), and D(0, 4). a. Prove that ABCD is a parallelogram. b. Prove that the diagonals of ABCD are perpendicular. c. Is ABCD a rhombus? Justify your answer. 19. If a diagonal separates a quadrilateral KLMN into two equilateral triangles, prove that KLMN is a rhombus with the measure of one angle equal to 60 degrees. 20. Prove that the diagonals of a rhombus separate the rhombus into four congruent right triangles. 21. Prove that if the diagonals of a quadrilateral are the perpendicular bisectors of each other, the quadrilateral is a rhombus. 22. Prove that if the diagonals of a parallelogram bisect the angles of the parallelogram, then the parallelogram is a rhombus. 23. Let P be any point on diagonal BD of rhombus ABCD. Prove that AP > CP. 24. The vertices of quadrilateral ABCD are A(2, 1), B(2, 4), C(2, 1), D(2, 4). a. Prove that ABCD is a parallelogram. b. Find the coordinates of E, the |
point of intersection of the diagonals. C PD A B c. Prove that the diagonals are perpendicular. d. Is ABCD a rhombus? Justify your answer. 25. ABCD is a parallelogram. The midpoint of AB is M, the midpoint of CD is N, and AM AD. a. Prove that AMND is a rhombus. b. Prove that MBCN is a rhombus. c. Prove that AMND is congruent to MBCN. Hands-On Activity In this activity, you will construct a rhombus given the diagonal. You may use geometry software or compass and straightedge. 1. First draw a segment measuring 12 centimeters. This will be the diagonal of the rhombus. The endpoints of this segment are opposite vertices of the rhombus. 2. Now construct the perpendicular bisector of this segment. 3. Show that you can choose any point on that perpendicular bisector as a third vertex of the rhombus. 4. How can you determine the fourth vertex of the rhombus? 5. Compare the rhombus you constructed with rhombuses constructed by your classmates. How are they alike? How are they different? 14365C10.pgs 7/10/07 8:50 AM Page 399 10-6 THE SQUARE DEFINITION A square is a rectangle that has two congruent consecutive sides. The Square 399 If consecutive sides and ), then rectangle ABCD is a square. AD AB if AB > AD of rectangle ABCD are congruent (that is, Theorem 10.17 A square is an equilateral quadrilateral. Given ABCD is a square with AB > BC. Prove AB > BC > CD > DA Proof A square is a rectangle and a rectangle is a parallelo- gram, so ABCD is a parallelogram. It is given that AB > BC gruent, so tive property of congruence,. Opposite sides of a parallelogram are conAB > CD BC > DA. Using the transiAB > BC > CD > DA. and D A C B Theorem 10.18 A square is a rhombus. Given Square ABCD Prove ABCD is a rhombus. Proof A square is an equilateral quadrilateral. If a quadrilateral is equilateral, then it is a rhombus. Therefore, ABCD is a rh |
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