Plane Geometry - Bruce E. Shapiro

112 SECTION 23. ANGLE-SIDE-ANGLEFigure 23.2: Illustration of proof of ASA (theorem 23.1).Since ∠ABC ∼ = ∠DEF (given) and ∠DEF ∼ = ∠ABG (by (23.1)), weconclude that ∠ABC ∼ = ∠ABG.By the angle construction postulate the rays −→ −→BG = BC (because theycorrespond to the same angle measured from −→ BA).Hence point G ∈ −→−→←→BC, and we see that ray BC intersects line AC at bothC and G. Hence line ←→ BC intersects ←→ AC at these same two points.But distinct non-parallel lines can only intersect in one point, so C = G.Hence △ABC ∼ = △ABG ∼ = △DEF .Corollary 23.2 (Converse of Isosceles Triangle Theorem) If △ABCis a triangle that satisfies ∠ACB ∼ = ∠ABC then AB = AC.Proof. Since BC ∼ = CB then by SAS △ABC ∼ = △BAC. Hence AB ∼ =AC.Theorem 23.3 Congruent Triangle Construction Theorem. It ispossible to construct a congruent copy of any triangle on a base, i.e., forany triangle △ABC, if DE ∼ = AB and H is either halfplane of ←→ DE thenthere exists a unique point F ∈ H such that △ABC ∼ = △DEF .Proof. Existence. Suppose that DE ∼ = AB, and choose a side H of AB.Let G ∈ H be chosen so that ∠ABC = ∠DEG (angle construction postulate).Choose F ∈ −→ EG such that EF = BC (ruler postulate).Hence by SAS, △ABC ∼ = △DEF« CC BY-NC-ND 3.0. Revised: 18 Nov 2012