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Theorem 4.2.5 (Hypotenuse-Leg Theorem). If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the two triangles are congruent. Restate ment IE A A BC and A DEE ara two right triangles with right anales at the vertices

Theorem 4.2.5 (Hypotenuse-Leg Theorem). If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the two triangles are congruent. Restate ment IE A A BC and A DEE ara two right triangles with right anales at the vertices

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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FIGURE 4.7: There is no Angle-Side-Side Theorem There is one significant special case in which Angle-Side-Side does hold; that is the case in which the given angle is a right angle. The theorem is known as the Hypotenuse-Leg Theorem. Definition 4.2.4. A triangle is a right triangle if one of the interior angles is a right angle. The side opposite the right angle is called the hypotenuse and the two sides adjacent to the right angle are called legs. B FIGURE 4.8: AABC is a right triangle; AB is the hypotenuse; AC and BC are legs Theorem 4.2.5 (Hypotenuse-Leg Theorem). If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the two triangles are congruent. Restatement. If AABC and ADEF are two right triangles with right angles at the vertices C and F, respectively, AB E DE, and BC EF, then AABC = ADEF. Proof. Exercise 4. Side-Side-Side is a also a valid triangle congruence condition. Its proof requires the following result, which is roughly equivalent to Euclid's Proposition 7. 76 Chapter 4 Neutral Geometry Theorem 4.2.6. If AABC is a triangle, DE is a segment such that DE = AB, and H is a half-plane bounded by DE, then there is a unique point F E H such that ADEF = AABC. Proof. Exercise 5. Theorem 4.2.7 (SSS). If AABC and ADEF are two triangles such that AB = DE, BC = EF, and CA = FD, then AABC = ADEF. Proof. Let AABC and ADEF be two triangles such that AB = DE, BC = EF and CA = FD (hypothesis). We must prove that AABC = ADEF. There exists a point G, on the opposite of AB from C, such that AABG = ADEF (Theorem 4.2.6). Since C and G are on opposite sides of AB, there is a point H such that H is between C and G and H lies on AB (Plane Separation Postulate). It is possible A or H = B. If not, then exactly one of the following will hold: H * A * B, or A * H * B, or A * B * H (Corollary 5.7.3). Thus there are five possibilities for the location of H relative to A and B. We will consider the various cases separately. that H = Н H=A В A В G FIGURE 4.9: Case 1, H = A, and Case 3, A * H * B Case 1. H = A. In this case ZAC B ZGCB and LAGB ZCGB. Since BC BG, we || have that LGCB = LCGB (Isosceles Triangle Theorem). Thus AABC = AABG (SAS). = B. The proof of this case is similar to the proof of Case 1. Case 2. H Case 3 A H J R Since H is between 4 and R H is in the interior of / ACR and in
Transcribed Image Text:FIGURE 4.7: There is no Angle-Side-Side Theorem There is one significant special case in which Angle-Side-Side does hold; that is the case in which the given angle is a right angle. The theorem is known as the Hypotenuse-Leg Theorem. Definition 4.2.4. A triangle is a right triangle if one of the interior angles is a right angle. The side opposite the right angle is called the hypotenuse and the two sides adjacent to the right angle are called legs. B FIGURE 4.8: AABC is a right triangle; AB is the hypotenuse; AC and BC are legs Theorem 4.2.5 (Hypotenuse-Leg Theorem). If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the two triangles are congruent. Restatement. If AABC and ADEF are two right triangles with right angles at the vertices C and F, respectively, AB E DE, and BC EF, then AABC = ADEF. Proof. Exercise 4. Side-Side-Side is a also a valid triangle congruence condition. Its proof requires the following result, which is roughly equivalent to Euclid's Proposition 7. 76 Chapter 4 Neutral Geometry Theorem 4.2.6. If AABC is a triangle, DE is a segment such that DE = AB, and H is a half-plane bounded by DE, then there is a unique point F E H such that ADEF = AABC. Proof. Exercise 5. Theorem 4.2.7 (SSS). If AABC and ADEF are two triangles such that AB = DE, BC = EF, and CA = FD, then AABC = ADEF. Proof. Let AABC and ADEF be two triangles such that AB = DE, BC = EF and CA = FD (hypothesis). We must prove that AABC = ADEF. There exists a point G, on the opposite of AB from C, such that AABG = ADEF (Theorem 4.2.6). Since C and G are on opposite sides of AB, there is a point H such that H is between C and G and H lies on AB (Plane Separation Postulate). It is possible A or H = B. If not, then exactly one of the following will hold: H * A * B, or A * H * B, or A * B * H (Corollary 5.7.3). Thus there are five possibilities for the location of H relative to A and B. We will consider the various cases separately. that H = Н H=A В A В G FIGURE 4.9: Case 1, H = A, and Case 3, A * H * B Case 1. H = A. In this case ZAC B ZGCB and LAGB ZCGB. Since BC BG, we || have that LGCB = LCGB (Isosceles Triangle Theorem). Thus AABC = AABG (SAS). = B. The proof of this case is similar to the proof of Case 1. Case 2. H Case 3 A H J R Since H is between 4 and R H is in the interior of / ACR and in
2:42 ll 5GE Gerard A. Venema Foundations of Geometry 2011 PDF - 3 MB Side-Side-Side is Euclid's Proposition 8. Euclid's proof is somewhat clumsy, relying on a rather convoluted lemma (Proposition 7). Heath attributes the more elegant proof, above, to Philo and Proclus (see [22], page 263). SSS can also be proved as a corollary of he Hinge Theorem (see Exercise 4.3.5), but that proof is not considered to be as elegant as the proof given here. Before leaving the subject of triangle congruence conditions we should mention that n neutral geometry the three angles of a triangle do not determine the triangle. This may seem obvious to you since you remember from high school that similar triangles are isually not congruent. However, that is a special situation in Euclidean geometry. One of the surprising results we will prove in hyperbolic geometry is that (in that context) Angle-Angle-Angle is a valid triangle congruence condition! ES 4.2 1. Prove the Converse to the Isosceles Triangle Theorem (Theorem 4.2.2). 2. Prove the Angle-Angle-Side Triangle Congruence Condition (Theorem 4.2.3). 3. Suppose AABC and ADEF are two triangles such that ZBAC = LEDF, AC = and CB DF, = FE (the hypotheses of ASS). Prove that either LABC and ZDEF are congruent or they are supplements. 4. Prove the Hypotenuse-Leg Theorem (Theorem 4.2.5). 5. Prove that it is possible to construct a congruent copy of a triangle on a given base (Theorem 4.2.6). REE INEQUALITIES FOR TRIANGLES The Exterior Angle Theorem gives one inequality that is always satisfied by the measures of the angles of a triangle. In this section we prove three additional inequalities that will be iseful in our study of triangles. The first of these theorems, the Scalene Inequality, extends he Isosceles Triangle Theorem and its converse. It combines Euclid's Propositions 18 and 19. The word scalene means "unequal" or “uneven." A scalene triangle is a triangle hat has sides of three different lengths. Theorem 4.3.1 (Scalene Inequality). In any triangle, the greater side lies opposite the greater angle and the greater angle lies opposite the greater side. Restatement. Let AABC be a triangle. Then AB > BC if and only if µ(LACB) > e(LBAC). Proof. Let A, B, and C be three noncollinear points (hypothesis). We will first assume he hypothesis AB > BC and prove that µ(LACB) > µ(LBAC). Since AB > BC, there exists a point D between A and B such that BD = BC (Ruler Postulate). D В FIGURE 4.11: The greater side lies opposite greater angle Now u(LACB) > u(LDCB) (Protractor Postulate, Part 4, and Theorem 3.3.10) and LDCB = LCDB (Isosceles Triangle Theorem). But LCDB is an exterior angle for
Transcribed Image Text:2:42 ll 5GE Gerard A. Venema Foundations of Geometry 2011 PDF - 3 MB Side-Side-Side is Euclid's Proposition 8. Euclid's proof is somewhat clumsy, relying on a rather convoluted lemma (Proposition 7). Heath attributes the more elegant proof, above, to Philo and Proclus (see [22], page 263). SSS can also be proved as a corollary of he Hinge Theorem (see Exercise 4.3.5), but that proof is not considered to be as elegant as the proof given here. Before leaving the subject of triangle congruence conditions we should mention that n neutral geometry the three angles of a triangle do not determine the triangle. This may seem obvious to you since you remember from high school that similar triangles are isually not congruent. However, that is a special situation in Euclidean geometry. One of the surprising results we will prove in hyperbolic geometry is that (in that context) Angle-Angle-Angle is a valid triangle congruence condition! ES 4.2 1. Prove the Converse to the Isosceles Triangle Theorem (Theorem 4.2.2). 2. Prove the Angle-Angle-Side Triangle Congruence Condition (Theorem 4.2.3). 3. Suppose AABC and ADEF are two triangles such that ZBAC = LEDF, AC = and CB DF, = FE (the hypotheses of ASS). Prove that either LABC and ZDEF are congruent or they are supplements. 4. Prove the Hypotenuse-Leg Theorem (Theorem 4.2.5). 5. Prove that it is possible to construct a congruent copy of a triangle on a given base (Theorem 4.2.6). REE INEQUALITIES FOR TRIANGLES The Exterior Angle Theorem gives one inequality that is always satisfied by the measures of the angles of a triangle. In this section we prove three additional inequalities that will be iseful in our study of triangles. The first of these theorems, the Scalene Inequality, extends he Isosceles Triangle Theorem and its converse. It combines Euclid's Propositions 18 and 19. The word scalene means "unequal" or “uneven." A scalene triangle is a triangle hat has sides of three different lengths. Theorem 4.3.1 (Scalene Inequality). In any triangle, the greater side lies opposite the greater angle and the greater angle lies opposite the greater side. Restatement. Let AABC be a triangle. Then AB > BC if and only if µ(LACB) > e(LBAC). Proof. Let A, B, and C be three noncollinear points (hypothesis). We will first assume he hypothesis AB > BC and prove that µ(LACB) > µ(LBAC). Since AB > BC, there exists a point D between A and B such that BD = BC (Ruler Postulate). D В FIGURE 4.11: The greater side lies opposite greater angle Now u(LACB) > u(LDCB) (Protractor Postulate, Part 4, and Theorem 3.3.10) and LDCB = LCDB (Isosceles Triangle Theorem). But LCDB is an exterior angle for
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