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1. Prove: If one interior angle of a triangle is right or obtuse, then both the other interior angles are acute.

1. Prove: If one interior angle of a triangle is right or obtuse, then both the other interior angles are acute.

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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must reject the RAA hypothesis and conclude that no such line m' exists. More on terminology. The proof above lays out a step-by-step process in which we apply the various postulates we have assumed, along with the theorems we have already proved, to verify that the perpendicular line must exist. When we present such a proof, we often say that we have “constructed" the perpendicular. What we mean by this is simply that we have shown that that existence of the perpendicular follows from the axioms. This use of the word “construction'" is quite different from what Euclid meant by a construction. Constructions in the sense of Euclid will be studied in Chapter 9. In the proof above, we proved unique- ness by showing that the existence of a second perpendicular line would lead to a contradic- tion. A slightly different strategy for proving uniqueness is to show that any line through P that is perpendicular to l must in fact be equal to m. Here is how that proof would be formulated: Assume m' is a line such that P P m m' lies on m' and m' 1 l. Let Q' be the point at which m' intersects l (which exists by definition of perpendicular). If Q' + Q, we would have a triangle APQQ' whose properties contradict the Exterior Angle Theorem, just as before. Hence it must be the case that Q' therefore m = m' by the Incidence Postulate. While the overall strategy for this second formulation of the proof is different, the essential proof is exactly the same as in the first formulation. A common error is to think that the Angle Construction Postulate implies uniqueness directly, but that is only the case if P lies on l. FIGURE 4.5: Two perpendiculars from P Q and EXERCISES 4.1 1. Prove: If one interior angle of a triangle is right or obtuse, then both the other interior angles are acute. 74 Chapter 4 Neutral Geometry 2. Let AABC be the spherical triangle shown in Figure 4.3. Perform the construction in the proof of Theorem 4.1.2 on the sphere, starting with this triangle. Convince yourself that the point F constructed is in the interior of LACD if and only if u(BAC) < 90°. Draw diagrams illustrating both possibilities. 4.2 TRIANGLE CONGRUENCE CONDITIONS The statement that two triangles are congruent means that the three interior angles of the first triangle are congruent to the corresponding angles in the second and that the three sides of the first triangle are congruent to the corresponding sides of the second. The Side-Angle-Side Postulate indicates that it is sometimes possible to conclude all six of these congruences from only three of them. SAS is just the first of several similar results, which are known as triangle congruence conditions. In this section we build on the Side-Angle-Side Postulate to prove the other familiar triangle congruence conditions. We begin with the Angle-Side-Angle triangle congruence condition. It is one half of Euclid's Proposition 26; the other half (Angle-Angle-Side) will be left as an exercise. Theorem 4.2.1 (ASA). If two angles and the included side of one triangle are congruent to the corresponding parts of a second triangle, then the two triangles are congruent. Restatement. IfAABC andADEF are two triangles such that LCAB = LFDE, AB = DE, and LABC = LDEF, then AABC = ADEF. Proof. Let AABC and ADEF be two triangles such that ZCAB = LFDE, AB = DE, and ZABC = LDEF (hypothesis). We must show that AABC = ADEF. A В D E FIGURE 4.6: One possible location for C' in proof of ASA There exists a point C' on AČ such that AC' = DF (Point Construction Postulate). Now AABC' = ADEF (SAS) and so LABC' = LDEF (definition of congruent triangles). Since LABC = LDEF (hypothesis), we can conclude that ZABC = LABC'. Hence BĆ = BC' (Protractor Postulate, Part 3). But BĆ can only intersect AC in at most one point (Theorem 3.1.7), so C = C' and the proof is complete. Angle-Side-Angle can be used to prove the converse to the Isosceles Triangle Theorem, which is Euclid's Proposition 6. Theorem 4.2.2 (Converse to the Isosceles Triangle Theorem). If AABC is a triangle such that LABC = LACB, then AB = AC. Proof. Exercise 1. Theorem 4.2.3 (AAS). If AABC and ADEF are two triangles such that LABC = LDEF, ZBCA = LEFD, and AC = DF, then AABC = ADEF.
Transcribed Image Text:must reject the RAA hypothesis and conclude that no such line m' exists. More on terminology. The proof above lays out a step-by-step process in which we apply the various postulates we have assumed, along with the theorems we have already proved, to verify that the perpendicular line must exist. When we present such a proof, we often say that we have “constructed" the perpendicular. What we mean by this is simply that we have shown that that existence of the perpendicular follows from the axioms. This use of the word “construction'" is quite different from what Euclid meant by a construction. Constructions in the sense of Euclid will be studied in Chapter 9. In the proof above, we proved unique- ness by showing that the existence of a second perpendicular line would lead to a contradic- tion. A slightly different strategy for proving uniqueness is to show that any line through P that is perpendicular to l must in fact be equal to m. Here is how that proof would be formulated: Assume m' is a line such that P P m m' lies on m' and m' 1 l. Let Q' be the point at which m' intersects l (which exists by definition of perpendicular). If Q' + Q, we would have a triangle APQQ' whose properties contradict the Exterior Angle Theorem, just as before. Hence it must be the case that Q' therefore m = m' by the Incidence Postulate. While the overall strategy for this second formulation of the proof is different, the essential proof is exactly the same as in the first formulation. A common error is to think that the Angle Construction Postulate implies uniqueness directly, but that is only the case if P lies on l. FIGURE 4.5: Two perpendiculars from P Q and EXERCISES 4.1 1. Prove: If one interior angle of a triangle is right or obtuse, then both the other interior angles are acute. 74 Chapter 4 Neutral Geometry 2. Let AABC be the spherical triangle shown in Figure 4.3. Perform the construction in the proof of Theorem 4.1.2 on the sphere, starting with this triangle. Convince yourself that the point F constructed is in the interior of LACD if and only if u(BAC) < 90°. Draw diagrams illustrating both possibilities. 4.2 TRIANGLE CONGRUENCE CONDITIONS The statement that two triangles are congruent means that the three interior angles of the first triangle are congruent to the corresponding angles in the second and that the three sides of the first triangle are congruent to the corresponding sides of the second. The Side-Angle-Side Postulate indicates that it is sometimes possible to conclude all six of these congruences from only three of them. SAS is just the first of several similar results, which are known as triangle congruence conditions. In this section we build on the Side-Angle-Side Postulate to prove the other familiar triangle congruence conditions. We begin with the Angle-Side-Angle triangle congruence condition. It is one half of Euclid's Proposition 26; the other half (Angle-Angle-Side) will be left as an exercise. Theorem 4.2.1 (ASA). If two angles and the included side of one triangle are congruent to the corresponding parts of a second triangle, then the two triangles are congruent. Restatement. IfAABC andADEF are two triangles such that LCAB = LFDE, AB = DE, and LABC = LDEF, then AABC = ADEF. Proof. Let AABC and ADEF be two triangles such that ZCAB = LFDE, AB = DE, and ZABC = LDEF (hypothesis). We must show that AABC = ADEF. A В D E FIGURE 4.6: One possible location for C' in proof of ASA There exists a point C' on AČ such that AC' = DF (Point Construction Postulate). Now AABC' = ADEF (SAS) and so LABC' = LDEF (definition of congruent triangles). Since LABC = LDEF (hypothesis), we can conclude that ZABC = LABC'. Hence BĆ = BC' (Protractor Postulate, Part 3). But BĆ can only intersect AC in at most one point (Theorem 3.1.7), so C = C' and the proof is complete. Angle-Side-Angle can be used to prove the converse to the Isosceles Triangle Theorem, which is Euclid's Proposition 6. Theorem 4.2.2 (Converse to the Isosceles Triangle Theorem). If AABC is a triangle such that LABC = LACB, then AB = AC. Proof. Exercise 1. Theorem 4.2.3 (AAS). If AABC and ADEF are two triangles such that LABC = LDEF, ZBCA = LEFD, and AC = DF, then AABC = ADEF.
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