Question 1 (SAS).Case 4.Н* А * В.The proof of this case isThis timevery much like the proof of Case 3.A is in the interior of LHCB and ZHGB (see Fig-ure 4.10). Thus µ(LACB) :and u(LAGB)tor Postulate, Part 4). But µ(LACH) = µ(LAGH)and µ(LHCB) = µ(LHCB) (Isosceles Triangle The-orem). Thus ZACB = LAGB (subtraction) and soΔΑΒC ΔΑΒG (SAS ) .µ(LHCB)u(LHGB) – e(LHGA) (Protrac-- μ(LHCA).%3DHCase 5. A * B * H. The proof of this case is similar tothe proof of Case 4.In each of the five cases we have proved thatFIGURE 4.10: Case 4, H * A * BAABC = AABG. But AABG = ADEF, so the proof is complete.Section 4.3Three inequalities for triangles 77Side-Side-Side is Euclid's Proposition 8. Euclid's proof is somewhat clumsy, relyingon 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 ofthe Hinge Theorem (see Exercise 4.3.5), but that proof is not considered to be as elegantas the proof given here.Before leaving the subject of triangle congruence conditions we should mention thatin neutral geometry the three angles of a triangle do not determine the triangle. Thismay seem obvious to you since you remember from high school that similar triangles areusually not congruent. However, that is a special situation in Euclidean geometry. Oneof the surprising results we will prove in hyperbolic geometry is that (in that context)Angle-Angle-Angle is a valid triangle congruence condition!EXERCISES 4.21. 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 = DF,and CB = FE (the hypotheses of ASS). Prove that either ZABC and ZDEF arecongruent 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).4.3 THREE INEQUALITIES FOR TRIANGLESThe Exterior Angle Theorem gives one inequality that is always satisfied by the measuresof the angles of a triangle. In this section we prove three additional inequalities that will beuseful in our study of triangles. The first of these theorems, the Scalene Inequality, extendsthe Isosceles Triangle Theorem and its converse. It combines Euclid's Propositions 18and 19. The word scalene means "unequal" or “uneven." A scalene triangle is a trianglethat has sides of three different lengths.Theorem 4.3.1 (Scalene Inequality). In any triangle, the greater side lies opposite thegreater 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) u(LBAC).Proof. Let A, B, and C be three noncollinear points (hypothesis). We will first assumethe hypothesis AB BC and prove that µ(LACB) µ(LBAC). Since AB BC, thereexists a point D between A and B such that BD = BC (Ruler Postulate).ABFIGURE 4.11: The greater side lies opposite greater angleNow µ(LACB) µ(LDCB) (Protractor Postulate, Part 4, and Theorem 3.3.10)and ZDCB = ZCDB (Isosceles Triangle Theorem). But LCDB is an exterior angle for78Chapter 4Neutral GeometryAADC (see Figure 4.11), so µ(LCDB) µ(LCAB) (Exterior Angle Theorem). Theconclusion follows from those inequalities.The proof of the converse is left as an exercise (Exercise 1).The second inequality is the familiar Triangle Inequality. It is Euclid's Proposition 20.Theorem 4.3.2 (Triangle Inequality). If A, B, and C are three noncollinear points, thenАС

Question

Question 1 (SAS).Case 4.Н* А * В.The proof of this case isThis timevery much like the proof of Case 3.A is in the interior of LHCB and ZHGB (see Fig-ure 4.10). Thus µ(LACB) :and u(LAGB)tor Postulate, Part 4). But µ(LACH) = µ(LAGH)and µ(LHCB) = µ(LHCB) (Isosceles Triangle The-orem). Thus ZACB = LAGB (subtraction) and soΔΑΒC ΔΑΒG (SAS ) .µ(LHCB)u(LHGB) – e(LHGA) (Protrac-- μ(LHCA).%3DHCase 5. A * B * H. The proof of this case is similar tothe proof of Case 4.In each of the five cases we have proved thatFIGURE 4.10: Case 4, H * A * BAABC = AABG. But AABG = ADEF, so the proof is complete.Section 4.3Three inequalities for triangles 77Side-Side-Side is Euclid's Proposition 8. Euclid's proof is somewhat clumsy, relyingon 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 ofthe Hinge Theorem (see Exercise 4.3.5), but that proof is not considered to be as elegantas the proof given here.Before leaving the subject of triangle congruence conditions we should mention thatin neutral geometry the three angles of a triangle do not determine the triangle. Thismay seem obvious to you since you remember from high school that similar triangles areusually not congruent. However, that is a special situation in Euclidean geometry. Oneof the surprising results we will prove in hyperbolic geometry is that (in that context)Angle-Angle-Angle is a valid triangle congruence condition!EXERCISES 4.21. 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 = DF,and CB = FE (the hypotheses of ASS). Prove that either ZABC and ZDEF arecongruent 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).4.3 THREE INEQUALITIES FOR TRIANGLESThe Exterior Angle Theorem gives one inequality that is always satisfied by the measuresof the angles of a triangle. In this section we prove three additional inequalities that will beuseful in our study of triangles. The first of these theorems, the Scalene Inequality, extendsthe Isosceles Triangle Theorem and its converse. It combines Euclid's Propositions 18and 19. The word scalene means "unequal" or “uneven." A scalene triangle is a trianglethat has sides of three different lengths.Theorem 4.3.1 (Scalene Inequality). In any triangle, the greater side lies opposite thegreater 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) >u(LBAC).Proof. Let A, B, and C be three noncollinear points (hypothesis). We will first assumethe hypothesis AB > BC and prove that µ(LACB) > µ(LBAC). Since AB > BC, thereexists a point D between A and B such that BD = BC (Ruler Postulate).ABFIGURE 4.11: The greater side lies opposite greater angleNow µ(LACB) > µ(LDCB) (Protractor Postulate, Part 4, and Theorem 3.3.10)and ZDCB = ZCDB (Isosceles Triangle Theorem). But LCDB is an exterior angle for78Chapter 4Neutral GeometryAADC (see Figure 4.11), so µ(LCDB) > µ(LCAB) (Exterior Angle Theorem). Theconclusion follows from those inequalities.The proof of the converse is left as an exercise (Exercise 1).The second inequality is the familiar Triangle Inequality. It is Euclid's Proposition 20.Theorem 4.3.2 (Triangle Inequality). If A, B, and C are three noncollinear points, thenАС < АВ + BС.Proof. Exercise 2. 74Chapter 4Neutral Geometry2. Let AABC be the spherical triangle shown in Figure 4.3. Perform the construction inthe proof of Theorem 4.1.2 on the sphere, starting with this triangle. Convince yourselfthat the point F constructed is in the interior of LACD if and only if u(LBAC) < 90°.Draw diagrams illustrating both possibilities.4.2 TRIANGLE CONGRUENCE CONDITIONSThe statement that two triangles are congruent means that the three interior angles ofthe first triangle are congruent to the corresponding angles in the second and that thethree 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 sixof these congruences from only three of them. SAS is just the first of several similarresults, which are known as triangle congruence conditions. In this section we build on theSide-Angle-Side Postulate to prove the other familiar triangle congruence conditions. Webegin with the Angle-Side-Angle triangle congruence condition. It is one half of Euclid'sProposition 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 tothe corresponding parts of a second triangle, then the two triangles are congruent.Restatement. IfAABC and ADEF are two triangles such that ZCAB = LFDE, AB = DE,and LABC = LDEF, then AABC = ADEF.Proof. Let AABC and ADEF be two triangles such that ZCAB = LFDE, AB = DE,and LABC = LDEF (hypothesis). We must show that AABC = ADEF.AВ DEFIGURE 4.6: One possible location for C' in proof of ASAThere 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 ZABC = LDEF (hypothesis), we can conclude that ZABC = LABC'. HenceB = BC' (Protractor Postulate, Part 3). But BC can only intersect AC in at most onepoint (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 TriangleTheorem, which is Euclid’s Proposition 6.Theorem 4.2.2 (Converse to the Isosceles Triangle Theorem). If AABC is a triangle suchthat LA BC 2 LACB, then АВ — АС.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.Section 4.2Triangle congruence conditions 75Proof. Exercise 2.The two conditions ASA and AAS show that any two angles plus a side determinethe entire triangle. Two sides plus one angle may or may not determine the triangle.If the angle is included between the two sides, then SAS implies that the triangle isdetermined. If the angle is not included between the sides, then the triangle is notcompletely determined and therefore ASS is not a valid congruence condition. Thefailure of Angle-Side-Side is illustrated in Figure 4.7. Even though the figure makes itclear that ASS is not a valid triangle congruence condition, two triangles that satisfy thehypotheses of Angle-Side-Side must be closely related-the relationship between them isspecified in Exercise 3.CAВAВFIGURE 4.7: There is no Angle-Side-Side TheoremThere is one significant special case in which Angle-Side-Side does hold; that is thecase in which the given angle is a right angle. The theorem is known as the Hypotenuse-LegTheorem.

Accepted Answer

We have to prove the converse of the isosceles triangle theorem

EXERCISES 4.2 1. Prove the Converse to the Isosceles Triangle Theorem (Theorem 4.2.2).