Answer:
Proved below by AAA similarity theorem.
Step-by-step 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)
