Plane Geometry - Bruce E. Shapiro

SECTION 26. SIDE-SIDE-SIDE 123By construction BG = EF ; by assumption EF = BC, hence BG = BC.Therefore △BCG is isosceles, and by the isosceles triangle theorem (theorem21.4)m(∠BCH) = γ = δ = m(∠BGH)Hencem(∠BCA) = ɛ = ζ = m(∠BGA) = m(∠EF D)so by SAS, △ACB ∼ = △AGB. Since △AGB ∼ = △DEF by construction,we have △ABC ∼ = △DEF .Case 4 (H ∗ A ∗ B) (see figure 26.3).Figure 26.3: Proof of SSS when H ∗ A ∗ B.H ∗ A ∗ B =⇒ −−→ CH ∗ −→ CA ∗ −→ CB =⇒ A is interior to ∠BCH.Similarly, H ∗ A ∗ B =⇒ −−→ GH ∗ −→ GA ∗ −→ GB =⇒ A is interior to ∠BGH.By the angle addition postulate,m(∠BCA) = m(∠BCH) − m(∠ACH)m(∠BGA) = m(∠BGH) − m(∠AGH)By construction AG = DF ; by assumption DF = AC; hence AC = AG.This means that △ACG is isosceles and hence by the isosceles triangletheorem (theorem 21.4), m(∠ACH) = m(∠AGH).By construction BG = EF ; by assumption EF = BC; hence BC = BGand △BCG is also isosceles. By the isosceles triangle theorem (theorem21.4), m(∠BCH) = m(∠BGH)Hencem(∠BCA) = m(∠BCH) − m(∠ACH)= m(∠BGH) − m(∠AGH)= m(∠BGA)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.