- Proof Exercise 2 And Close The Reason For The Name Is The Fact That The Lengths Of The Two Sides Of The Triangle The 1 (133.11 KiB) Viewed 52 times
Proof. Exercise 2. and close, The reason for the name is the fact that the lengths of the two sides of the triangle, The
-
- Site Admin
- Posts: 899603
- Joined: Mon Aug 02, 2021 8:13 am
Proof. Exercise 2. and close, The reason for the name is the fact that the lengths of the two sides of the triangle, The
Proof. Exercise 2. and close, The reason for the name is the fact that the lengths of the two sides of the triangle, The third inequality is Euclid's Proposition 24. It is known as the much like a door hinge. The theorem is a generalization of SAS and is sometimes called fixed but the angle is allowed to vary. This means that the triangle can open the SAS Inequality. A B D FIGURE 4.12: Hinge Theorem: If μ(LBAC) < (LEDF), then BC < EF E Theorem 4.3.3 (Hinge Theorem). If AABC and ADEF are two triangles such that AB = DE, AC = DF, and (LBAC) < (LEDF), then BC < EF. ← Proof. Let AABC and ADEF be two triangles such that AB = DE, AC = (LBAC) < (LEDF). We must show that BC < EF. Find a point G, on the same side DF, and of AB as C, such that ABG = ADEF (Theorem 4.2.6). The proof will be complete if we show that BG > BC. Since C is in the interior of LBAG (Theorem 3.4.5), AC must intersect BG in a C, then C is between B and G and the conclusion BGBC follows immediately. Hence we may assume for the remaindon point J (Crossbar Theorem). If J =