# Theorem 6.5

Last updated at Nov. 27, 2017 by Teachoo

Last updated at Nov. 27, 2017 by Teachoo

Transcript

Theorem 6.5 (SAS Similarity) If one angle of a triangle is equal to one angle of the other triangle and sides including these angles are proportional then the triangles are similar. Given :- Two triangles ∆ABC and ∆DEF such that ∠A = ∠D ABDE = ACDF To Prove : ∆ABC ~ ∆DEF Construction :- Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ. Proof :- Given ABDE = CAFD And DP = AB, DQ = AC ∴ DPDE = DQDF DEDP = DFDQ Subtracting 1 on both sides DEDP − 1 = DFDQ − 1 DF − DPDP = DF − DQDQ PEDP = QFDQ DPPE = DQQF Using theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side. ∴ PQ ∥ EF. Thus, ∠P = ∠E and ∠Q = ∠F In ∆ABC and ∆DPQ AB = DP AC = DQ BC = PQ ⇒ ∆ABC ≅ ∆DPQ ∴ ∠B = ∠P ∠C = ∠Q But From (1) ∠P = ∠E and ∠Q = ∠F Therefore, ∠B = ∠P = ∠E and ∠C = ∠Q = ∠F Therefore, In Δ ABC & Δ DEF ∠B = ∠E ∠C = ∠F ∴ ∆ABC ~ ∆DEF Hence Proved

Chapter 6 Class 10 Triangles

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .