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Theorem 14.19. In an affine plane, two distinct lines can meet in at most one point. | Proof. Let \( \ell \) and \( m \) be distinct lines in an affine plane. If \( \ell \parallel m \), then they have no points in common and the proof is completed. Suppose \( \ell \nparallel m \) . For the sake of reaching a contradiction assume that they meet at more than one point, say \( P \) and \( Q \) . This implie... | Yes |
Theorem 14.20. There are at least two points on every line in any affine plane. | Proof. Let \( \ell \) be a line in an affine plane. Let points \( A, B, C \) and \( D \) be the four points given by Axiom A-3. If any two of these points are on \( \ell \), then the proof is complete. Otherwise, at least three of these points are not on \( \ell \) . Without loss of generality, suppose \( A \) is not o... | Yes |
Theorem 14.21. In an affine plane, for any line \( \ell \) there is a point not on \( \ell \) . | Proof. Let \( \ell \) be an arbitrary line. By Theorem 14.17, let \( m \) be a line distinct from \( \ell \) . By Theorem 14.19, lines \( \ell \) and \( m \) have at most one point in common. By Theorem 14.20, there is at least one point on \( m \) that is not on \( \ell \) . | Yes |
Theorem 14.22 [Proclus’ Axiom]. In an affine plane, if a line (distinct from two different parallel lines) meets one of the two parallel lines, then it meets the other. | Proof. Suppose there are three distinct lines, \( \ell, m \) and \( k \), such that \( m\parallel k \), and \( \ell \) meets \( m \) at point \( P \) . We must show that \( \ell \) meets \( k \) . For the sake of an eventual contradiction, suppose \( \ell \parallel k \) . Thus, there are two lines through \( P \) that ... | Yes |
Theorem 14.23 [Euclid I.30]. In an affine plane, two distinct lines, each parallel to a third line, must be parallel to each other. | Proof. Suppose there are three distinct lines, \( \ell, m \) and \( k \), such that \( \ell \parallel k \) and \( m\parallel k \) . We must show that \( \ell \parallel m \) . For the sake of an eventual contradiction, suppose \( \ell \) meets \( m \) at point \( P \) . Thus, there are two lines through \( P \) that are... | Yes |
Lemma 14.24. In a finite affine plane, if point \( P \) is not on line \( \ell \), then the number of lines on point \( P \) is one more than the number of points on \( \ell \) . | Proof. By Axiom A-2, there is exactly one line on point \( P \), say \( m \), that is parallel to \( \ell \) . Therefore, every line on \( P \) other than \( m \) must meet \( \ell \) . Combining this with Theorem 14.19, every line on \( P \) other than \( m \) meets \( \ell \) in exactly one point. By Axiom A-1, there... | Yes |
Theorem 14.25. In a finite affine plane, each line has the same number of points. | Proof. Let points \( A, B, C \) and \( D \) be the four points given by Axiom A-3. By Axiom A-3, \( A \) is not on the distinct lines \( {BC} \) or \( {CD} \) . By Lemma 14.24, the number of lines on \( A \) is one more than the number of points on \( {BC} \) . Likewise, the number of lines on \( A \) is one more than ... | Yes |
Theorem 14.28. In a finite affine plane of order \( n \), there are \( {n}^{2} \) points. | Proof. Let \( \ell \) be a line in the plane of order \( n \) . Since the order of the plane is \( n,\ell \) consists of exactly \( n \) distinct points, say \( {B}_{1},{B}_{2},\ldots ,{B}_{n} \) . By Theorem 14.21, let \( P \) be a point not on \( \ell \) . By Axiom A-1, there exists unique line \( {B}_{1}P \) . Notic... | Yes |
Theorem 14.30. In a finite affine plane of order \( n \), there are \( {n}^{2} + n \) lines. | Proof. Let \( P \) be a point in the plane. By Corollary 14.27, \( P \) is on \( n + 1 \) distinct lines, say \( {\ell }_{1},{\ell }_{2},\ldots ,{\ell }_{n + 1} \) . Each of these lines forms a parallel class consisting of \( n \) distinct lines. These parallel classes are mutually disjoint, meaning no line can be in m... | Yes |
Theorem 15.4. In Neutral geometry, every isometry transforms straight lines onto straight lines. | Proof. Let \( f \) be an isometry and let \( \ell = \overrightarrow{AB} \) be a line. To show that \( f\left( \ell \right) \) is a line, we need to show two things. First, if \( C \) is another point on line \( \ell \), we must show \( {C}^{\prime } = f\left( C\right) \) lies on the unique line joining \( {A}^{\prime }... | Yes |
Lemma 15.5. In Neutral geometry, if an isometry fixes two points, then it fixes the line joining them. | Proof. Let \( f \) be an isometry and let \( A \) and \( B \) be two points of the plane such that \( f\left( A\right) = A \) and \( f\left( B\right) = B \) . By Theorem 15.4, \( f \) maps line \( \overrightarrow{AB} \) to itself.\n\nMoreover, if \( C \) is on line \( \overrightarrow{AB} \) and \( f\left( C\right) = {C... | Yes |
Theorem 15.9. In Neutral geometry, if an isometry fixes three noncollinear points, then it is the identity. | Proof. Suppose that \( f \) is an isometry such that \( f\left( A\right) = A, f\left( B\right) = B \) and \( f\left( C\right) = C \) , where \( A, B \) and \( C \) are noncollinear. From Lemma 15.5, we know that \( f \) fixes lines \( \overrightarrow{AB} \) , \( \overrightarrow{BC} \) and \( \overrightarrow{AC} \) . Le... | Yes |
Corollary 15.10. If two isometries agree at three noncollinear points, then they agree everywhere. | Proof. Let \( A, B \), and \( C \) be three noncollinear points, and let \( f \) and \( g \) be isometries such that \( f\left( A\right) = g\left( A\right), f\left( B\right) = g\left( B\right) \) and \( f\left( C\right) = g\left( C\right) \) . Consider the isometry \( {g}^{-1} \circ f \) . Notice that, for example, \( ... | Yes |
Lemma 15.17. Let \( {T}_{AB} \) be a translation. If \( C \) and \( D \) are any two points in the plane with \( {C}^{\prime } = {T}_{AB}\left( C\right) \) and \( {D}^{\prime } = {T}_{AB}\left( D\right) \), and, if \( {C}^{\prime } \) is not collinear with \( C \) and \( D \), then \( C{C}^{\prime }{D}^{\prime }D \) is... | Proof. Let \( {T}_{AB} \) be a translation, and let \( C \) and \( D \) be any two points in the plane with \( {C}^{\prime } = {T}_{AB}\left( C\right) \) and \( {D}^{\prime } = {T}_{AB}\left( D\right) \) where \( {C}^{\prime } \) is not collinear with \( C \) and \( D \) . By definition, \( C{C}^{\prime } = {AB} = D{D}... | Yes |
Theorem 15.19. If \( A, B \) and \( C \) are any three points, then\n\n\[ \n{T}_{BC} \circ {T}_{AB} = {T}_{AC} \n\] | From this theorem, we have that \( {T}_{AB} \circ {T}_{BA} = {T}_{AA} = i \) . Thus,\n\n\[ \n{T}_{AB}^{-1} = {T}_{BA} \n\] | No |
Consider the composition \( {T}_{DB} \circ {T}_{AC} \), where points \( A, B, C \) and \( D \) are given by \( ▱{ABCD} \), as shown in Figure 15.13. To identify the composition \( {T}_{DB} \circ {T}_{AC} \) as a single translation, we first determine the intermediate image \( {T}_{AC}\left( {▱{ABCD}}\right) = \) \( ▱{A... | Here, we mean each point \( x \) is mapped to \( {x}^{\prime } \) . For example, since we translate by \( \overrightarrow{AC} \), we have \( {A}^{\prime } = {T}_{AC}\left( A\right) = C \) . Next, we find the final image of the resulting composition, \( {T}_{DB}\left( {{T}_{AC}\left( {▱{ABCD}}\right) }\right) = {T}_{DB}... | Yes |
Theorem 15.26. Let \( {R}_{A,\alpha } \) and \( {R}_{B,\beta } \) be two rotations with \( 0 \leq \alpha ,\beta < {2\pi } \) . The composition \( {R}_{B,\beta } \circ {R}_{A,\alpha } \) is either a rotation or a translation. | Proof. If \( A = B \), then \( {R}_{B,\beta } \circ {R}_{A,\alpha } = {R}_{A,\alpha + \beta } \), and we are done.\n\nAssume \( A \neq B \) . Let \( \ell = \overrightarrow{AB} \) . Construct a line \( m \) through \( A \) such that the directed angle from \( m \) to \( \ell \) at \( A \) is \( \alpha /2 \) . Construct ... | No |
Theorem 15.27. Let \( {R}_{A,\alpha } \) be a nonzero rotation and let \( T \) be a translation. The compositions \( {R}_{A,\alpha } \circ T \) and \( T \circ {R}_{A,\alpha } \) are both rotations with angle \( \alpha \) . | Proof. Let \( {R}_{A,\alpha } \) be a nonzero rotation and let \( T \) be a translation where \( T\left( B\right) = A \) . We will show that \( {R}_{A,\alpha } \circ T \) is a rotation with angle \( \alpha \) . Since \( B = {T}^{-1}\left( A\right) \), we have \( T = {T}_{BA} \) . Let \( C \) be the midpoint of \( {BA} ... | No |
Theorem 15.29. Let \( {T}_{AB} \) be a translation and \( {F}_{\ell } \) be a reflection. Then \( {F}_{\ell } \circ {T}_{AB} \) and \( {T}_{AB} \circ {F}_{\ell } \) are both reflections or both glide reflections. | Proof. We will prove that \( {F}_{\ell } \circ {T}_{AB} \) is either a reflection or a glide reflection and leave the proof that \( {T}_{AB} \circ {F}_{\ell } \) is the same type of isometry to the reader.\n\nCase 1. Suppose \( {AB} \) lies on \( \ell \) . Then \( {F}_{\ell } = {F}_{AB} \), and combining this with Exer... | No |
Theorem 15.30. If \( {R}_{A,\alpha } \) is a rotation and \( {F}_{\ell } \) a reflection, then \( {F}_{\ell } \circ {R}_{A,\alpha } \) and \( {R}_{A,\alpha } \circ {F}_{\ell } \) are both reflections or both glide reflections. | Proof. We will prove that \( {F}_{\ell } \circ {R}_{A,\alpha } \) is either a reflection or a glide reflection and leave the proof that \( {R}_{A,\alpha } \circ {F}_{\ell } \) is the same type of isometry to the reader.\n\n, a radius \( r \), and a point \( C \), where \( C \) neither lies on the circle of inversion nor is \( A \), to find \( {C}^{\prime } = {I}_{A, r}\left( C\right) \) . | Proof.\n\nCase 1. Suppose \( C \) lies inside the circle centered at \( A \) . Join \( {AC} \) and extend it to form ray \( \overrightarrow{AC} \) . Using Euclid I.11, construct the perpendicular to \( {AC} \) at \( C \) . Let \( B \) be its intersection with the circle. Join \( {AB} \) . Using Euclid I.11, construct t... | No |
Lemma 15.35. Suppose \( {I}_{A, r} \) is an inversion, and \( A, B \) and \( C \) are distinct, noncollinear points. If \( {B}^{\prime } = {I}_{A, r}\left( B\right) \) and \( {C}^{\prime } = {I}_{A, r}\left( C\right) \), then \( \bigtriangleup {ABC} \sim \bigtriangleup A{C}^{\prime }{B}^{\prime } \) . | Proof. Consider triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup A{C}^{\prime }{B}^{\prime } \) . By definition, these triangles share \( \angle A \) . Moreover, since \( {BA} \cdot {B}^{\prime }A = {CA} \cdot {C}^{\prime }A = {r}^{2} \), we have\n\n\[ \frac{{B}^{\prime }A}{{C}^{\prime }A} = \frac{CA}{BA} \]\... | Yes |
Theorem 15.36. Given center \( A \) and radius \( r \), the inversion \( {I}_{A, r} \) maps\n\n- a line that passes through \( A \) onto itself,\n\n- a line that does not pass through \( A \) onto a circle that passes through \( A \) ,\n\n- a circle that passes through \( A \) onto a line that does not pass through \( ... | Proof.\n\nCase 1. Assume that \( \ell \) is a line that passes through \( A \), and let \( C \) be a point on \( \ell \) other than \( A \) . By definition, \( {C}^{\prime } \) lies on the ray \( \overrightarrow{AC} \), and hence the inversion \( {I}_{A, r} \) maps \( \ell \) onto itself.\n\n![40fcc26d-fa4b-4fe4-b639-2... | Yes |
Theorem 15.39. Let \( {I}_{A, r} \) be an inversion, and \( c \) be a circle that intersects the circle of inversion.\n\n- If \( c \) and the circle of inversion are orthogonal, then \( {I}_{A, r} \) maps \( c \) onto itself.\n\n- If \( c \) contains both a point \( D \) and its image \( {I}_{A, r}\left( D\right) = E \... | Proof. Suppose that \( c \) and the circle of inversion intersect at a point \( B \) .\n\nCase 1. Assume that circle \( c \) and the circle of inversion are orthogonal. By Euclid III. \( {16},{AB} \) is tangent to circle \( c \) . Let \( D \) be a point on \( c \) that does not also lie on the circle of inversion. The ... | Yes |
Lemma 15.41. If \( \ell \) and \( m \) are ultraparallel lines, then \( H{F}_{m} \circ H{F}_{\ell } \) is a hyperbolic translation. | Proof. Assume \( \ell \) and \( m \) are ultraparallel lines. By Theorem 13.34, \( \ell \) and \( m \) have a unique common perpendicular \( \overrightarrow{EF} \), where \( E \) lies on \( \ell \) and \( F \) lies on \( m \), as illustrated in Figure 15.47. Let \( G \) lie on \( \overrightarrow{EF} \) such that \( F \... | Yes |
Theorem 15.46. If the hyperbolic line, \( \ell \), is represented by a vertical ray, then the hyperbolic reflection across \( \ell \) is the same as the Euclidean reflection across \( \ell \), namely \( H{F}_{\ell } = {F}_{\ell } \) . | Proof. Given hyperbolic line \( \ell \) with equation \( x = c \), consider a point \( A = \left( {{x}_{1},{x}_{2}}\right) \) , not on \( \ell \) . Let \( {A}^{\prime } = {F}_{\ell }\left( A\right) \) be the image of \( A \) under the Euclidean reflection across \( \ell \) . Then \( {A}^{\prime } = \left( {{x}_{2},{y}_... | No |
Theorem 15.47. The Euclidean inversion \( {I}_{C, k} \), where \( C = \left( {c,0}\right) \), preserves the arc length of a path \( \gamma \) in the Poincaré Half-plane model. | Proof. Any path \( \gamma \) can be parameterized as \( x\left( t\right) = c + r\left( t\right) \cos t \) and \( y\left( t\right) = r\left( t\right) \sin t \) for some \( r\left( t\right) \) where \( \alpha \leq t \leq \beta \) . [Note that these are merely modified polar coordinates centered at \( C = \left( {c,0}\rig... | Yes |
Theorem 15.48. Consider a semicircle centered at \( C = \left( {c,0}\right) \) with radius \( k \) . The Euclidean inversion \( {I}_{C, k} \) maps\n\n- a vertical ray that passes through \( C \) onto itself,\n\n- a vertical ray that does not pass through \( C \) onto a semicircle centered on the \( x \) -axis that pass... | Proof. We first note that the image of point \( A \) under the inversion \( {I}_{C, k} \) lies on the ray \( \overrightarrow{CA} \) . Thus \( {I}_{C, k} \) maps the upper half-plane onto itself. We consider the following four cases:\n\nCase 1. Suppose that \( \ell \) is the vertical ray \( x = c \) . Then by Theorem 15... | Yes |
Lemma 15.49. Suppose the hyperbolic line \( \ell \) is represented by a semicircle centered at \( C = \left( {c,0}\right) \) with radius \( k \) . Consider a point \( A \) not on \( \ell \), and let \( {A}^{\prime } = {I}_{C, k}\left( A\right) \) . If \( m \) is the unique perpendicular hyperbolic line from \( A \) to ... | Proof.\n\nCase 1. Suppose \( A \) lies on the vertical ray \( x = c \) . Then the unique perpendicular from \( A \) to \( \ell \) is the vertical ray \( x = c \) . Since \( {A}^{\prime } \) lies on the ray \( \overrightarrow{CA},{A}^{\prime } \) also lies on \( x = c \) .\n\nCase 2. Suppose \( A \) does not lie on \( x... | Yes |
Theorem 15.50. If the hyperbolic line, \( \ell \), is represented by a semicircle centered at \( C = \) \( \left( {c,0}\right) \) with radius \( k \), then the hyperbolic reflection across \( \ell \) is the Euclidean inversion in \( \ell \) , that is, \( H{F}_{\ell } = {I}_{C, k} \) . | Proof. Consider a point \( A \) that does not lie on our hyperbolic line, \( \ell \) . Let \( {A}^{\prime } = {I}_{C, k}\left( A\right) \) . By Lemma 15.49, \( \ell \bot A{A}^{\prime } \) . Let \( B \) be the intersection of \( \ell \) and \( A{A}^{\prime } \) . We note that \( {I}_{C, k} \) fixes \( B \) . Since \( {I... | Yes |
Theorem 16.1. Given a unit length and constructible lengths \( a \) and \( b \), we can construct the following:\n\n(1) \( a + b \)\n\n(2) \( a - b \) (assuming \( a > b \) )\n\n(3) \( {ab} \)\n\n(4) \( \frac{a}{b} \)\n\n(5) \( \sqrt{a} \). | Proof. We leave the constructions of \( a + b \) and \( a - b \) to the reader.\n\nConstruction of \( {ab} \) : Consider a ray starting at a point \( A \) . Construct \( {AB} = 1 \), and on the same ray, construct \( {AC} = a \) . On another ray starting at \( A \), construct \( {AD} = b \) . Join \( {BD} \) . Construc... | No |
Lemma 16.7 [Lines]. If a line ax \( + {by} + c = 0 \) passes through points \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \) , where \( {x}_{1},{x}_{2},{y}_{1} \) and \( {y}_{2} \) are all surds, then the coefficients \( a, b \) and \( c \) are also surds. | Proof. Let a line pass through the points \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \), where \( {x}_{1},{x}_{2},{y}_{1} \) and \( {y}_{2} \) are all surds. We leave it to the reader to show that the equation of the line is given by \( \left( {{y}_{2} - {y}_{1}}\right) x + \left( {{x}... | No |
Lemma 16.8 [Circles]. If a circle \( {x}^{2} + {y}^{2} + {Ax} + {By} + C = 0 \) has center; \( \left( {h, k}\right) \), and radius, \( r \), where \( h, k \) and \( r \) are surds, then \( A, B \) and \( C \) are also surds. | Proof. A circle with center, \( \left( {h, k}\right) \), and radius \( r \), is given by the equation \( {\left( x - h\right) }^{2} + \) \( {\left( y - k\right) }^{2} = {r}^{2} \) . Simplifying, we get \( {x}^{2} + {y}^{2} + \left( {-{2h}}\right) x + \left( {-{2k}}\right) y + \left( {{h}^{2} + {k}^{2} - {r}^{2}}\right)... | Yes |
Lemma 16.9 [Line meets Line]. If the nonparallel lines \( {ax} + {by} + c = 0 \) and \( {a}^{\prime }x + \) \( {b}^{\prime }y + {c}^{\prime } = 0 \) have surd coefficients, then their point of intersection has surd coordinates. | Proof. We leave it to the reader to show that the intersection of the lines \( {ax} + {by} + c = 0 \) and \( {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \) is the point\n\n\[ \left( {\frac{b{c}^{\prime } - {b}^{\prime }c}{a{b}^{\prime } - {a}^{\prime }b},\frac{{a}^{\prime }c - a{c}^{\prime }}{a{b}^{\prime } - {... | No |
Lemma 16.10 [Line meets Circle]. If the line with equation \( {ax} + {by} + c = 0 \) intersects the circle with equation \( {x}^{2} + {y}^{2} + {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \), and both have surd coefficients, then any point intersection has surd coordinates. | Proof. We leave it to the reader to show that any intersection point of the line, \( {ax} + \) \( {by} + c = 0 \), and the circle, \( {x}^{2} + {y}^{2} + {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \), is of the form \( \left( {x, y}\right) \) where\n\n\[ x = \frac{-B \pm \sqrt{{B}^{2} - {4AC}}}{2A} \]\n\nwith ... | No |
Lemma 16.11 [Circle meets Circle]. If two intersecting circles, with equations \( {x}^{2} + {y}^{2} + {ax} + {by} + c = 0 \) and \( {x}^{2} + {y}^{2} + {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \), both have surd coefficients, then any point of intersection has surd coordinates. | Proof. We leave it to the reader to show the intersection of the circles, \( {x}^{2} + {y}^{2} + {ax} + {by} + c = 0 \) and \( {x}^{2} + {y}^{2} + {a}^{\prime }x + {b}^{\prime }y + {c}^{\prime } = 0 \), is of the form \( \left( {x, y}\right) \) where\n\n\[ x = \frac{-B \pm \sqrt{{B}^{2} - {4AC}}}{2A} \]\n\nwith \( A = ... | No |
Lemma 16.12 [Length of Segment]. If a segment joins the points \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \) , where \( {x}_{1},{x}_{2},{y}_{1} \) and \( {y}_{2} \) are all surds, then the length of the segment is also a surd. | Proof. The length of the segment joining the points \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \) is\n\n\[ \sqrt{{\left( {x}_{1} - {x}_{2}\right) }^{2} + {\left( {y}_{1} - {y}_{2}\right) }^{2}}. \]\n\nThus, any constructible line segment has a surd length, that is, it can be expressed ... | Yes |
Lemma 11.13. A regular \( n \) -gon is constructible if and only if its central angle \( \frac{{360}^{ \circ }}{n} \) is constructible. | If we could construct a regular 9-gon, then we could construct its central angle, a \( {40}^{ \circ } \) angle. Bisecting this, we would be able to construct a \( {20}^{ \circ } \) angle, which we know is impossible. Thus, we cannot construct a regular 9-gon. | No |
Theorem 1.1.1. (Euclid I-32) In Euclidean geometry the angle sum of a triangle is \( {180}^{ \circ } \) . | Proof. Let \( \bigtriangleup {ABC} \) be a triangle and construct the line \( \overset{\overleftrightarrow{} }{DE} \) parallel to \( \overline{BC} \) through \( A \) (Figure 1.1). Then (by Euclid’s proposition I-29) \( \angle {DAB} \cong \angle {CBA} \) and \( \angle {CAE} \cong \angle {ACB} \) . Thus the angles of the... | Yes |
Theorem 1.1.2. (Euclid I-47) In a Euclidean right triangle the square on the hypotenuse has the same area as the squares on the sides: that is, \( {a}^{2} + {b}^{2} = {c}^{2} \) . | ## Proof. (See Exercises 1.1.17 and 1.1.18.) - | No |
Theorem 1.1.3. No rational number equals \( \sqrt{2} \) . (The diagonal of a square is incommensurable with its side.) | Proof. Suppose, for a contradiction, that there were two integers \( p \) and \( q \) such that \( p/q = \sqrt{2} \) . Without loss of generality, assume that \( p \) and \( q \) are not both even. Otherwise we could factor out any common factors of 2 . Then \( {\left( p/q\right) }^{2} = 2 \), or \( {p}^{2} = 2{q}^{2} ... | Yes |
Given segment \( \overline{AB} \), construct its perpendicular bisector. | Solution. By postulate 3, we may draw circle \( {C}_{1} \) with center \( A \) and radius \( {AB} \) and circle \( {C}_{2} \) with center \( B \) and radius \( {AB} \) . (See Figure 1.11.) Let \( D \) and \( E \) be the intersections of the circles. Use postulate 1 to draw \( \overline{DE} \) . We claim that \( \overli... | Yes |
Construct a square if you know one side of it, say, \( \overline{AB} \) (Euclid I-46). | Solution. Construct circle \( {C}_{1} \) with center \( B \) and radius \( {AB} \) . Use postulate 2 to extend \( \overline{AB} \) to \( Z \) on \( {C}_{1} \) . (See Figure 1.14.) Note that \( B \) is the midpoint of \( \overline{AZ} \) . Use Example 1 to construct the perpendicular \( \overline{BD} \), where \( D \) i... | Yes |
Example 3. Construct a regular pentagon given one side of it, \( \overline{PQ} \) . | Solution. Given the segment \( \overline{PQ} \), we use Example 2 to construct a square \( {ABCD} \) with \( \overline{AB} \cong \) \( \overline{PQ} \) and Figure 1.15 to construct \( \overline{AG} \) . By Exercise 1.2.3 the length \( {AG} \) equals \( {PQ}\frac{1 + \sqrt{5}}{2} \) . The Pythagorean theorem (I-47) show... | Yes |
Example 4. Construct the tangents to a circle from an outside point. | Solution. Let \( P \) be outside the circle with center \( C \) (Fig. 1.17). Construct the midpoint \( M \) of \( \overline{CP} \) and the circle with center \( M \) and radius \( \overline{MC} \) . The circle intersects the original circle in two points, \( A \) and \( B \) . We claim that \( \overline{PA} \) and \( \... | No |
Using a line \( k \) and a point \( P \) not on \( k \), construct a line \( m \) parallel to \( k \) with \( P \) on \( m \) . | Solution. Let \( k \) be the line through points \( Q \) and \( R \) and \( P \) be a point not on \( k \) . (See Figure 1.35.) Construct the circles \( {C}_{1} \) with center \( Q \) and radius \( {QR},{C}_{2} \) with center \( P \) and radius \( {PR},{C}_{3} \) with center \( P \) and radius \( {QR} \), and \( {C}_{4... | Yes |
Theorem 1.3.1. Euclid's fifth postulate is equivalent to Playfair's axiom, assuming that Euclid's first 28 propositions hold. | Proof. (Euclid \( \Rightarrow \) Playfair) Suppose that the fifth postulate holds and that we are given a point \( P \) not on a line \( k \) . Example 4 gives us one parallel, say, \( m \), where \( \overline{PQ} \) is perpendicular to both \( m \) and \( k \), as shown in Figure 1.37. For Playfair’s axiom we need to ... | Yes |
Theorem 1.3.2. (part of I-29) Suppose \( \overrightarrow{AB}\parallel \overrightarrow{CD} \) and \( A \) and \( D \) are on opposite sides of \( \overrightarrow{BC} \) . Then the opposite interior angles are congruent: \( \angle {ABC} \cong \angle {BCD} \) . | Proof. Suppose \( \overset{\overleftrightarrow{} }{AB}\parallel \overset{\overleftrightarrow{} }{CD} \) and \( A \) and \( D \) are on opposite sides of \( \overset{\overleftrightarrow{} }{BC} \) . Following Example 1, we can construct \( \overrightarrow{CE} \) with \( E \) on the same side of \( \overrightarrow{BC} \)... | Yes |
Theorem 1.3.3. If \( {ABCD} \) is a parallelogram, then opposite sides are congruent: \( \overline{AB} \cong \overline{CD} \) and \( \overline{AD} \cong \overline{BC} \) . | Proof. Let \( {ABCD} \) be a parallelogram and construct \( \overline{AC} \) . (See Figure 1.39.) By the definition of a parallelogram, \( \overset{\overleftrightarrow{} }{AB}\parallel \overset{\overleftrightarrow{} }{CD} \) . Also, \( \angle {CAB} \) and \( \angle {ACD} \) are alternate interior angles, so by I-29, th... | Yes |
Show that the three triangles in Figure 1.46 are similar. | Solution. Each triangle is a right triangle by the converse of the Pythagorean theorem (Euclid I-48). Elementary trigonometry confirms that the other angles are congruent. Furthermore, the sides of the largest triangle \( \left( {{10},6\text{, and 8}}\right) \) and the smallest triangle \( \left( {6,{3.6}\text{, and 4.... | Yes |
Theorem 1.4.1. (Euclid VI-2) Let \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) be two rays with \( D \) between \( A \) and \( B \) and \( E \) between \( A \) and \( C \) . Then \( \overline{BC}\parallel \overline{DE} \) if and only if \( \bigtriangleup {ABC} \sim \bigtriangleup {ADE} \) . | Proof. Construct segments \( \overline{BE} \) and \( \overline{CD} \), as in Figure 1.47. \( \left( \Rightarrow \right) \) First suppose that \( \overline{BC}\parallel \overline{DE} \) . Then the corresponding angles are equal by I-29. Triangles \( \bigtriangleup {BDE} \) and \( \bigtriangleup {CED} \) have the same ar... | Yes |
Theorem 1.4.2. (Euclid VI-4) If two triangles have two pairs of corresponding angles congruent, then the triangles are similar. | Proof. Let \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) have \( \angle A \cong \angle {A}^{\prime } \) and \( \angle B \cong \angle {B}^{\prime } \) . By Theorem 1.1.1, \( \angle C \cong \angle {C}^{\prime } \) . We must show that the three pairs of sides are proportional.... | Yes |
Theorem 1.4.4. (Euclid VI-6) If two pairs of corresponding sides of two triangles are proportional and their included angles are congruent, then the triangles are similar. | Proof. Exercise 1.4.17. | No |
Theorem 1.4.5. If the lengths of the sides of \( \bigtriangleup {ABC} \) are \( k \) times the lengths of the corresponding sides of \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \), then the area of \( \bigtriangleup {ABC} \) is \( {k}^{2} \) times the area of \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C... | ## Proof. Exercise 14.18. | No |
Use calculus to find a formula for the volume of a pyramid. | Solution. Without loss of generality select a coordinate axis so that the origin is at the apex of the pyramid and the \( x \) -axis is perpendicular to the base (Figure 1.65). Suppose that the area of the base is \( B \) and that the \( x \) -coordinate of points on the base is \( h > 0 \), the height of the pyramid. ... | No |
Figure 1.66 illustrates how we can dissect a cube into three pyramids. A cube with a side of \( s \) is a prism with a base of area \( {s}^{2} \) and so a volume of \( {s}^{3} \) . Each of the square pyramids has a base with area \( {s}^{2} \) and a height of \( s \) . | So the sum of the volumes of the pyramids is \( 3\left( {\frac{1}{3}s \cdot {s}^{2}}\right) = {s}^{3} \) , as expected. In general all polyhedra can be dissected into pyramids, much as all polygons can be dissected into triangles. | Yes |
Theorem 1.5.1. Euler’s Formula If a convex polyhedron has \( V \) vertices, E edges, and \( F \) faces, then \( V - E + F = 2 \) . | Outline of Reasoning. Given a convex polyhedron (Figure 1.68) we can stretch the net of its vertices and edges to lay it out on a plane (Figure 1.69). The number of vertices and edges remains the same, but the number of faces is one less than the original polyhedron. If a face of the net is not a triangle, we divide it... | No |
Theorem 1.5.2. Descartes’ Formula If a convex polyhedron has \( V \) vertices, then the angle sum of all of the angles around all the vertices is \( {360}^{ \circ }\left( {V - 2}\right) \) . | Proof. We derive this result from Euler’s formula. We rewrite \( V - E + F = 2 \) as \( V - 2 = \) \( E - F \) . The left side multiplied by 360 is Descartes’ formula: \( {360}^{ \circ }\left( {V - 2}\right) = {360}^{ \circ }\left( {E - F}\right) = \) \( {180}^{ \circ }\left( {{2E} - {2F}}\right) \) . Next we relate th... | Yes |
Example 4. Find the edge lengths and angle measures for the two-frequency dome shown in Figure 1.72, if the radius \( {OA} \) of the sphere is 1. | Solution. From Exercise 1.5.15 we know that the length \( {AB} \approx {1.05146} \) . Because \( {D}^{\prime } \) and \( {E}^{\prime } \) are the midpoints of \( \overline{AB} \) and \( \overline{AC}, A{D}^{\prime } = {D}^{\prime }{E}^{\prime } \approx {0.52573} \) . By the Pythagorean theorem, \( O{D}^{\prime } \appro... | Yes |
Theorem 1.5.3. The area of a spherical triangle is proportional to the difference between its angle sum and \( \pi \) . More precisely, on a sphere with radius \( r \), the area of a spherical triangle with angle measures of \( \alpha ,\beta \), and \( \gamma \) is \( \left( {\alpha + \beta + \gamma - \pi }\right) {r}^... | Proof. The area covered by the lunes \( {BAC}{A}^{\prime },{ABC}{B}^{\prime } \), and \( {ACB}{C}^{\prime } \) is \( {2\pi }{r}^{2} \), or half the sphere because the opposite lunes cover a symmetric region. The three lunes each cover \( \bigtriangleup {ABC} \). Then\n\n\[ \n{2\pi }{r}^{2} = \operatorname{Area}\left( {... | Yes |
Claim: A rectangle inscribed in a square is a square. | Proof. Let the rectangle MNPQ be inscribed in the square ABCD (Figure 2.3). Drop perpendiculars from \( P \) to \( \overline{AB} \) and from \( Q \) to \( \overline{BC} \) at \( R \) and \( S \), respectively. Clearly, \( \overline{PR} \cong \overline{QS} \) because the segments match the sides of the square \( {ABCD} ... | Yes |
Theorem 2.1.1. For three points \( A, B \), and \( C \), if \( A\left( B\right) C \), then not \( C\left( A\right) B \), not \( B\left( C\right) A \), and not \( B\left( A\right) C \) . | Proof. Let \( A, B \), and \( C \) be any points and suppose \( A\left( B\right) C \) . By axiom (i), \( C\left( B\right) A \) and so by axiom (ii), not \( C\left( A\right) B \) . Next suppose for a contradiction that \( B\left( C\right) A \) was correct. By axiom (i) we would have \( A\left( C\right) B \), in violatio... | No |
Theorem 2.1.2. For three points \( A, B \), and \( C \), if \( A\left( B\right) C \), then \( A, B \), and \( C \) are distinct points. | Proof. Let \( A, B \), and \( C \) be any points and suppose \( A\left( B\right) C \) . By axiom (ii), \( A \neq C \) . Next suppose for a contradiction that \( B = C \) . From this equality and the given \( A\left( B\right) C \), we could switch \( B \) and \( C \) to get \( A\left( C\right) B \) . But that contradict... | No |
Theorem 2.1.3. There are at least four distinct points. | Proof. By axiom (iii) there are at least two distinct points, say, \( D \) and \( E \) . By axiom (iv) there is a point \( F \) so that \( D\left( E\right) F \) . Again by axiom (iv) there is a point \( G \) so that \( E\left( F\right) G \) . From Theorem 2 we know that \( D \neq E, D \neq F, E \neq F, E \neq G \), and... | Yes |
Theorem 2.1.4. If \( S \) and \( T \) are convex sets, then their intersection is convex. | ## Proof. See Exercise 2.1.5. \( \blacksquare \) | No |
We can stretch a string taut on the surface of a sphere to construct a spherical line segment. Similarly, we can construct a spherical circle by fixing one end of a taut string and swinging the other end around it. Experiment informally on a ball or sphere. Confirm that, within reason, Euclid's five postulates hold in ... | The construction does not hold in this model but does in the usual model, therefore it is independent of Euclid's postulates. For more on spherical geometry, see Section 1.5, Henderson [7], and McCleary [12]. | No |
Exercise 2.2.12 asked you to prove that there were at least four points on a line. You might have expected that problem to ask you to continue and prove that there were infinitely many points. But in fact we can't force more than seven points on a line from the axioms, as the following model shows. | By point we mean any of the seven vertices of a regular heptagon, which is what we interpret as a line. (See Figure 2.11.) Interpret a point \( Q \) to be between points \( P \) and \( R \) if and only if \( Q \) is on the perpendicular bisector of the Euclidean segment \( \overline{PR} \) . For example, in Figure 2.11... | Yes |
To show isomorphism, we need first to find a potential match of the points. As a first attempt, let’s try matching \( a \) and \( b \) in Figure 2.14 with \( A \) and \( C \), respectively, in Figure 2.6. We quickly run into trouble: \( a \) and \( b \) share a third dot on the same triangle, but \( A \) and \( C \) do... | For the next attempt, try matching \( a \) and \( b \) in Figure 2.14 with \( A \) and \( B \) in Figure 2.6. Do the choices restrict the remaining matches? Yes: only one other dot is on the triangle on \( a \) and \( b \) and only one other dot on the segment for \( A \) and \( B \) . So they must be matched. Further,... | Yes |
Show that the medians of a triangle intersect in the point two-thirds of the way from a vertex to the opposite midpoint. | Without loss of generality, pick the axes so that the vertices of the triangle shown in Figure 3.3 are \( \\left( {0,0}\\right) ,\\left( {a,0}\\right) \), and \( \\left( {b, c}\\right) \) . Verify that the midpoints of the sides are \( \\left( {\\frac{a}{2},0}\\right) ,\\left( {\\frac{b}{2},\\frac{c}{2}}\\right) \), an... | No |
Find the set (locus) of points \( P \) such that \( P \) is the center of a circle tangent to two given lines. | Solution. If the two lines are parallel, then a circle tangent to both must have its center on the line midway between them. In this case the midline is the locus. If the two lines intersect at a point \( Q \), then the centers of circles tangent to both lines must be on one of their two angle bisectors. Illustrate the... | No |
Example 2. Find an equation for an ellipse. | Solution. Without loss of generality, let the foci \( F \) and \( {F}^{\prime } \) have coordinates \( \left( {f,0}\right) \) and \( \left( {-f,0}\right) \) . Let \( P = \left( {x, y}\right) \) be a point on the ellipse. The definition of an ellipse gives the following equation.\n\n\[ \sqrt{{\left( x - f\right) }^{2} +... | Yes |
Example 3. An equation of a hyperbola is \( {x}^{2}/{a}^{2} - {y}^{2}/{b}^{2} = 1 \), and an equation of a parabola is \( y = a{x}^{2} \). | Solution. See Exercise 3.2.7. \( \diamond \) | No |
Use analytic geometry and calculus to verify that the method of Exercise 3.2.3 gives tangents to a parabola. | For computational ease and to match the equation \( y = a{x}^{2} \) for a parabola from Example 3, let the directrix \( m \) have equation \( y = - 1 \) and the focus \( F \) be \( \left( {0,1}\right) \), as in Figure 3.12. For a point \( W = \left( {w, - 1}\right) \) on \( m \), the midpoint of \( \overline{FW} \) is ... | Yes |
The hyperbola \( {x}^{2} - {y}^{2} = 1 \) is closely related to the degenerate conic \( {x}^{2} - {y}^{2} = 0 \), as illustrated in Figure 3.13. We can factor \( {x}^{2} - {y}^{2} = 0 \) into \( \left( {x - y}\right) \left( {x + y}\right) = 0 \) or the lines \( y = x \) and \( y = - x \) . They are called the asymptote... | We can use calculus to make this notion more precise. Write the hyperbola as \( y = \pm \sqrt{{x}^{2} - 1} \) . Then the expression \( \sqrt{{x}^{2} - 1} - x \) compares the difference between the \( y \) -coordinates of a point on the upper right branch of the hyperbola and the on \( y = x \) with the same \( x \) -co... | Yes |
If we pick \[ F = \left( {\frac{\sqrt{2}}{2},\frac{-\sqrt{2}}{2}}\right) ,\;{F}^{\prime } = \left( {\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}}\right) ,\;\text{ and }\;k = 2\sqrt{2}, \] as shown in Figure 3.14, the ellipse has the equation \( \frac{3}{4}{x}^{2} + \frac{1}{2}{xy} + \frac{3}{4}{y}^{2} = 1 \) . | Note that \( {ac} - {b}^{2} = \) \( \left( {3/4}\right) \left( {3/4}\right) - \left( {1/4}\right) \left( {1/4}\right) = 1/2 > 0 \) and that the determinant is \( - 1/2 \) . \( \diamond \) | No |
Use calculus and trigonometry to show the reflection property of the parabola: All rays from the focus are reflected by the parabola in rays parallel to its axis of symmetry. | Turn the parabola horizontally, as shown in Figure 3.15, and use calculus and trigonometry. Verify that the parabola \( y = \pm \sqrt{4kx} \) has focus \( F = \left( {k,0}\right) \) . For a point \( P = \left( {x,\sqrt{4kx}}\right) \) on the upper half of the parabola, verify that the slope of the line from \( F \) to ... | Yes |
For \( x\left( t\right) = {t}^{3} - t, y\left( t\right) = {t}^{2} - {t}^{4} \), and \( - {1.1} < t < {1.1} \), the point \( \left( {x\left( t\right), y\left( t\right) }\right) \) traces out the curve shown in Figure 3.19. We can think of the graph starting at the lower left for \( t = - {1.1} \), heading up to the righ... | Setting the derivative \( {y}^{\prime }\left( t\right) = {2t} - 4{t}^{3} \) equal to 0 shows that the curve reaches its maxima when \( t = \pm \sqrt{2}/2 \) and its local minimum when \( t = 0 \) . | Yes |
A cycloid is the curve traced by a point on the circumference of a circle as the circle rolls along a line. Find parametric equations for a cycloid for a circle of radius \( r \) . | Solution. We solve with \( r = 1 \) . Let a circle of radius 1 roll along the \( x \) -axis with the point starting at \( \left( {0,0}\right) \) when \( t = 0 \) . Figure 3.20 illustrates a general point on the cycloid. As the circle rolls, the center moves along the line \( y = 1 \) . When the circle has turned an ang... | Yes |
In 1694 Bernoulli investigated a curve he called a lemniscate, given by \( {r}^{2} = \) \( \cos \left( {2\theta }\right) \) whose graph appears in Figure 3.26. We need only consider angles where \( \cos \left( {2\theta }\right) \geq 0 \) . By the periodicity of the cosine, we can restrict the angles to just \( - \frac{... | Similarly, \( \theta = - \frac{\pi }{4} \) is tangent at the origin, which helps to explain the figure-eight shape of the curve. | Yes |
Find the parabola \( y = a{x}^{2} + {bx} + c \) through the points \( \left( {0,2}\right) ,\left( {1,3}\right) \), and \( \left( {3, - 7}\right) \). | When we replace \( x \) and \( y \) by the values for each point we get three first-degree equations in \( a, b \), and \( c : 2 = c,3 = a + b + c \), and \( - 7 = {9a} + {3b} + c \) . Solving the system gives \( a = - 2, b = 3 \), and \( c = 2 \) or \( y = - 2{x}^{2} + {3x} + 2 \) . | Yes |
Theorem 3.4.1. Given \( n + 1 \) points \( {P}_{j} = \left( {{x}_{j},{y}_{j}}\right) \), for \( j = 0 \) to \( n \), with no two \( {x}_{j} \) equal, there is exactly one nth degree polynomial \( y = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \) such that for each \( j \) , \( {y}_{j} = {a}_{0} + {a}_{1}{x}_{j} + \cd... | Proof. See Strang [15, 80]. | No |
Example 3. Verify that the slopes of the Bézier curve determined by the four points \( {P}_{0} = \) \( \left( {{b}_{0},{c}_{0}}\right) ,{P}_{1} = \left( {{b}_{1},{c}_{1}}\right) ,{P}_{2} = \left( {{b}_{2},{c}_{2}}\right) \), and \( {P}_{3} = \left( {{b}_{3},{c}_{3}}\right) \) match the slopes of the segments \( \overli... | Solution. The slope of \( \overline{{P}_{0}{P}_{1}} \) is \( \frac{{c}_{1} - {c}_{0}}{{b}_{1} - {b}_{0}} \) . The slope of the Bézier curve with points \( \left( {x\left( t\right), y\left( t\right) }\right) \) at \( t = w \) is \( \frac{{y}^{\prime }\left( w\right) }{{x}^{\prime }\left( w\right) } \) . Now \( {B}_{0}^{... | Yes |
In \( {\mathbb{R}}^{2} \), convert the formula for the line \( \alpha \overrightarrow{u} + \overrightarrow{v} \), where \( \overrightarrow{u} = \left( {2,3}\right) \) and \( \overrightarrow{v} = \left( {4,5}\right) \), to the more familiar equation of a line. | We can rewrite the formula \( \alpha \left( {2,3}\right) + \left( {4,5}\right) \) as \( \left( {{2\alpha } + 4,{3\alpha } + 5}\right) \) . That is, a point on the line has the property that when \( x = {2\alpha } + 4 \), we have \( y = {3\alpha } + 5 \) . Solve in the first equation to find \( \alpha = \frac{1}{2}x - 2... | Yes |
Use analytic geometry and linear algebra to verify Postulate 8 of the SMSG axioms: In three-dimensional geometry, if two distinct planes intersect, they intersect in a line. | Let’s begin with two distinct planes \( \alpha \overrightarrow{u} + \beta \overrightarrow{v} + \overrightarrow{w} \) and \( \gamma \overrightarrow{r} + \delta \overrightarrow{s} + \overrightarrow{t} \) . Then the coordinates of the six points are known, but the values of the four numbers \( \alpha ,\beta ,\gamma \), an... | Yes |
Given a point \( P \) not on a line \( k \), there are infinitely many lines on \( P \) that have no points on \( k \). | Proof. Let the two lines indicated by the characteristic axiom be \( l \) and \( m \). Pick points \( A \) on \( k \) and \( B \) and \( C \) on \( l \) such that line \( m \) enters \( \bigtriangleup {ABC} \) at \( P \). By Pasch’s axiom, \( m \) intersects the triangle on another side, without loss of generality at \... | Yes |
Theorem 4.2.2. If \( l \) and \( m \) are the two sensed parallels to \( k \) at \( P \), they have the same angle of parallelism. | Proof. As shown in Figure 4.11, let \( \overline{AP} \bot k \) and \( \overrightarrow{PB} \) and \( \overrightarrow{PC} \) be the sensed parallels to \( k \) at \( P \) . For a contradiction, let the angles of parallelism differ, say, \( m\angle {APB} < m\angle {APC} \) . Construct inside \( \angle {APC} \) a new angle... | Yes |
Corollary 4.2.3. All angles of parallelism are less than a right angle. Two lines with a common perpendicular are ultraparallel. | Proof. See Exercise 4.2.5. | No |
Theorem 4.2.4. Let \( P \) be a point not on \( k \) and let \( m \) be a sensed parallel to \( k \) at \( P \) . If \( S \) is any other point on \( m \), then \( m \) also is a sensed parallel to \( k \) at \( S \) . | Proof. Without loss of generality, let \( m \) be a right-sensed parallel to \( k \) at \( P \) .\n\nCase 1. Suppose that \( S \) is on \( m \) to the left of \( P \) and \( U \) is on \( m \) with \( S \) between \( P \) and \( U \) . Because \( m \) and \( k \) do not intersect, \( m \) is either the right-sensed par... | No |
Theorem 4.2.6. (Modified Pasch’s axiom for omega triangles) If a line \( k \) contains a point interior to \( \bigtriangleup {AB\Pi } \) and contains one of the vertices, then \( k \) intersects the opposite side of \( \bigtriangleup {AB\Pi } \) . | Proof. Let \( C \) be in \( \bigtriangleup {AB\Pi } \) . First consider the line \( k = \overrightarrow{AC} \) (Figure 4.17). Because \( \overrightarrow{A\Pi } \) is the sensed parallel to \( \overrightarrow{B\Pi } \) at \( A, k \) intersects \( \overrightarrow{B\Pi } \), say at \( D \) . The line \( \overrightarrow{BC... | Yes |
Theorem 4.2.7. (Euclid I-16 for omega triangles) The measure of an exterior angle of an omega triangle is greater than the measure of the opposite interior angle. | Proof. Let \( \bigtriangleup {AB\Pi } \) be an omega triangle and extend \( \overline{AB} \) (Figure 4.18). We prove \( \angle {CA\Pi } \) is greater than \( \angle {AB\Pi } \) by eliminating the other two possibilities through contradictions.\n\nCase 1. For a contradiction, assume that \( m\angle {CA\Pi } < m\angle {A... | Yes |
Theorem 4.2.8. In omega triangles \( \bigtriangleup {AB\Omega } \) and \( \bigtriangleup {CD\Lambda } \), if \( \overline{AB} \cong \overline{CD} \) and \( \angle {AB\Omega } \cong \) \( \angle {CD\Lambda } \), then \( \bigtriangleup {AB\Omega } \cong \bigtriangleup {CD\Lambda } \). | Proof. For a contradiction, assume that \( \angle {BA\Omega } \) is not congruent to \( \angle {DC\Lambda } \), and without loss of generality \( \angle {DC\Lambda } \) has a smaller measure. Construct \( \angle {BAP} \) inside \( \angle {BA\Omega } \) with \( \angle {BAP} \cong \) \( \angle {DC\Lambda } \) . Then \( \... | Yes |
Theorem 4.2.9. If corresponding angles of omega triangles are congruent, then the omega triangles are congruent. | ## Proof. See Exercise 4.2.11. ∎ | No |
Theorem 4.3.1. The summit angles of a Saccheri quadrilateral are congruent. The base and summit are perpendicular to the line on their midpoints. | Proof. Let \( \overline{AB} \) be the base of a Saccheri quadrilateral \( {ABCD} \) and \( \overline{AC} \) and \( \overline{BD} \) the diagonals (Figure 4.22). Then \( \bigtriangleup {ABC} \cong \bigtriangleup {BAD} \) by SAS, and by SSS \( \bigtriangleup {ADC} \cong \bigtriangleup {BCD} \) . Hence the summit angles, ... | Yes |
Theorem 4.3.2. The summit angles of a Saccheri quadrilateral are acute. | Proof. On a Saccheri quadrilateral \( {ABCD} \) with base \( \overline{AB} \), let \( \overrightarrow{C\Pi } \) and \( \overrightarrow{D\Pi } \) be left-sensed parallels to \( \overrightarrow{AB} \) (Figure 4.24). By Theorem 4.2.7, the exterior angle \( \angle {ED\Pi } \) of omega triangle \( \bigtriangleup {DC\Pi } \)... | Yes |
Theorem 4.3.3. The angle sum of a triangle is less than \( \pi \) . | Proof. In a triangle \( \bigtriangleup {ABC} \), let \( D \) be the midpoint of \( \overline{AB} \) and \( E \) be the midpoint of \( \overline{AC} \). Construct the perpendiculars \( \overline{AF},\overline{BG} \), and \( \overline{CH} \) to \( \overline{DE} \).\n\nWe claim that either \( D \) is between \( F \) and \... | No |
Corollary 4.3.4. The angle sum of a quadrilateral is less than \( {2\pi } \) . | Proof. In a quadrilateral \( {ABCD} \) draw \( \overline{BD} \) . If \( A \) and \( C \) are on opposite sides of \( \overrightarrow{BD} \), then \( {ABCD} \) is the union of \( \bigtriangleup {ABD} \) and \( \bigtriangleup {CBD} \), as in Figure 4.28. In this case, the angle sum of \( {ABCD} \) is the sum for the two ... | No |
Theorem 4.3.5. If two triangles have corresponding angles congruent, then the triangles are congruent. | Proof. Let \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) have corresponding angles congruent (Figure 4.29). If one pair of corresponding sides is congruent, then by ASA the triangles would be congruent. For a contradiction, assume without loss of generality that \( \overlin... | No |
Verify that a triangle and the associated Saccheri quadrilateral from Theorem 4.3.3 have the same area. | Solution. We consider case 1 of Theorem 4.3.3, as shown in Figure 4.25. In the proof we showed that \( \bigtriangleup {AFD} \cong \bigtriangleup {BGD} \) and \( \bigtriangleup {AEF} \cong \bigtriangleup {CEH} \) . The quadrilateral \( {BDEC} \) is congruent to itself. By postulate 18 the pieces of \( \bigtriangleup {AB... | No |
Theorem 4.4.2. If the polygons \( A \) and \( B \) are equivalent and the polygons \( B \) and \( C \) are equivalent, then \( A \) is equivalent to \( C \) . | Proof. Divide \( A \) and \( B \) into families of congruent triangles, \( {A}_{i} \) and \( {B}_{i} \) . Divide \( B \) and \( C \) into families of congruent triangles, \( {B}_{j}^{\prime } \) and \( {C}_{j}^{\prime } \), where \( {B}_{i} \) and \( {B}_{j}^{\prime } \) can be different (Figure 4.33). We subdivide the... | Yes |
Theorem 4.4.3. Two Saccheri quadrilaterals with congruent summits and congruent summit angles have congruent sides and bases and so are congruent. | Proof. Let \( {ABCD} \) and \( {EFGH} \) be two Saccheri quadrilaterals with \( \overline{CD} \cong \overline{GH},\angle {ADC} \cong \) \( \angle {EHG} \), and bases \( \overline{AB} \) and \( \overline{EF} \) . We need to show that their sides and bases are congruent. Without loss of generality suppose, for a contradi... | No |
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