In the given question we have been asked to prove that BAC is an isosceles triangle. If ∠D=∠E and AD/BD=AE/EC.
ML Aggarwal Avichal Publication Class 10, similarity question no 6
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AD/DB = AE/EC (Given)
Therefore, DE ∥ BC (Converse of Basic Proportionality Theorem) So, ∠D = ∠B and ∠E = ∠C (Corresponding angles) (1) But ∠D = ∠E (Given)
Therefore, ∠B = ∠C [ From (1)]
So, AB = AC (Sides opposite to equal angles)
i.e., BAC is an isosceles triangle