Question

Parallelogram ABCD is given. Draw line EF so that it goes through the vertex A. Point E lies on the side BC and point F lies on the extension of DC . Prove that △ABE∼△FCE.

Answer 1

Proved below by AAA similarity theorem.

Explanation:

Line EF is drawn through the vertex A of the parallelogram ABCD. E is on the side BC and F lies on the extension of DC.

Please refer to the attached figure.

In this figure,

In triangles Δ AEB and Δ CEF,

∠ AEB = ∠ CEF (vertically opposite angles)

CF is the extension of side DC.

Since DC || AB, DF || AB and FA is a transversal.

Therefore, ∠ BAE = ∠ EFC (alternate interior angles)

Also, DC || AB and CB is a transversal.

Therefore, ∠ ABE = ∠ FCE (alternate interior angles)

Hence, Δ ABE ~ Δ FCE (By AAA similarity criterian)