I came across a question from similarity chapter in which I found difficulty in solving the problem in which we are to prove that △ABE∼△CFB if E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
ML Aggarwal Avichal Publication class 10, similarity, question no 16a
In △ABE and △CFB,
∠ABE=∠CFB (Alternate angles)
∠BAE=∠BCF (opposite angles of a parallelogram)
∴ By AA criterion of similarity, △ABE ∼ △CFB