Q4 of 49 Page 45

In the adjoining figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that DE || BC. [ hint:-use the concept of alternate angles.]

In Δ AED and Δ ABC we have


BA = AD(Given)


CA = AE(Given)


∠EAD = ∠BAC (Vertically Opposite)


So Δ AED and Δ ABC are congruent by S.A.S. axiom of congruency


So we can say


∠AED = ∠ACB(Corresponding parts of Congruent triangles)


So ∠AED & ∠ACB forms a pair of alternate interior angles


Hence DE || BC


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