Plane Geometry - Bruce E. Shapiro

SECTION 35. SIMILAR TRIANGLES 179Then DE = r × AB andDFAC = r = EFBCHence DF = r × AC and EF = r × BCCorollary 35.4 (SAS Criterion for Similarity) If △ABC and △DEFare two triangles such that ∠CAB = ∠F DE and AB/AC = DE/DF then△ABC ∼ △DEF .Proof. Refer again to figure 35.1; suppose that α = δ and ABAC = DEDF .If AB = DE, the △ABC ∼ = △DEF and the theorem follows.Suppose that AB ≠ DE. We can assume without loss of generality thatAB > DE (otherwise we just relabel the two triangles).Pick B ′ such that AB ′ = DE and construct line m parallel to BC as before.As in the earlier proof, m intersects BC by Pasch’s theorem at some pointC ′ .Since B and C are on one side of m and A is on the other side, A ∗ C ′ ∗ C.By the corresponding angles theorem η = β and γ = ∠AC ′ B ′ . HenceBy the parallel projection theorem△ABC ∼ △AB ′ C ′ (35.1)AB ′AB = AC′AC ⇒ AB′AC ′ = ABAC = DEDFwhere we have used the hypotheses for the theorem in the last equality.But by construction AB ′ = DE, so thatDEAC ′ = DEDFBy SAS (DF = AC ′ , α = δ, AB ′ = DE),Substituting equation 35.1 gives⇒ DF = AC′△AB ′ C ′ ∼ = △DEF△ABC ∼ △DEFThe following theorem is also the converse of the Fundamental Theorem ofSimilar Triangles.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.