Plane Geometry - Bruce E. Shapiro

122 SECTION 26. SIDE-SIDE-SIDECase 1 (H = A) (see figure 26.1).By construction, △ABG ∼ = △DEF hence BG = EF ; by assumption EF =BC. Hence BC = BG.Thus triangle CGB is isosceles, and by the isosceles triangle theorem (theorem21.4), ∠BCA ∼ = ∠BGA.By congruence of triangles, ∠BGA ∼ = ∠EF D.Hence ∠BCA ∼ = ∠EF D.Since AG = DF = AC, and BG = EF = BC, by side angle side, △ABC ∼ =△ABG ∼ = △DEF .Case 2 (H = B). The proof is completely analogous to Case 1.Case 3 (A ∗ H ∗ B) (see figure 26.2).Figure 26.2: Proof of SSS when A ∗ H ∗ B.Since A ∗ H ∗ B, −→ CA ∗ −−→ CH ∗ −→ CB =⇒ that H is in the interior of ∠ACB(labeled as ɛ in fig. 26.2) (Betweenness for Rays, see 16.8).Similarly, −→ GA ∗ −−→ GH ∗ −→ GB =⇒ that H is in the interior of ∠BGA (labeledas ζ in fig. 26.2).By the protractor postulatem(∠ACB) = ɛ = α + γ = m(∠ACH) + m(∠BCH)m(∠(BGA) = ζ = β + δ = m(∠AGH) + m(∠BGH)By construction AG = DF ; by assumption DF = AC, hence AC = AG.Therefore △AGC is isosceles, and by the isosceles triangle theorem (theorem21.4)m(∠AGH) = β = α = m(∠ACH)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012