Plane Geometry - Bruce E. Shapiro

180 SECTION 35. SIMILAR TRIANGLESCorollary 35.5 (SSS Criterion for Simlarity) If △ABC and △DEFare two triangles such that AB/DE = AC/DF = BC/EF then △ABC ∼△DEF .Proof. Suppose AB/DE = AC/DF = BC/EF in triangles △ABC and△DEF .If AB = DE then AC = DF and BC = EF , so by SSS the triangles arecongruent.Suppose AB ≠ DE. Assume AB > DE (if not, relabel). Construct△AB ′ C ′ as in the proof of the previous theorem; as we saw in that proof△AB ′ C ′ ∼ △ABC.By the fundamental theorem of similar triangles,ABAB ′ = ACAC ′ = BCB ′ C ′ (35.2)Since AB ′ = DE by construction, and by hypothesis AB/DE = AC/DF ,ACDF = ABDE = ABAB ′ = ACAC ′and therefore AC ′ = DF . Also by hypothesis AB/DE = BC/EF in 35.2givesBCEF = ABDE = ACAC ′ = BCB ′ C ′hence B ′ C ′ = EF . Thus △AB ′ C ′ ∼ = △DEF by SSS. Since △AB ′ C ′ ∼△ABC, we have △ABC ∼ △DEF .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012