Answer :

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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