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E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\triangle ABE \sim \triangle CFB$.
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In $\triangle ABE$ and $$\begin{aligned}& \triangle CFB \\ & \angle EAB = \angle BCF \\ & \angle AEB = \angle CBF \\ & \Rightarrow \triangle ABE \sim \triangle CFB\end{aligned}$$_a_1.png)
In $\triangle ABE$ and $$\begin{aligned}& \triangle CFB \\ & \angle EAB = \angle BCF \\ & \angle AEB = \angle CBF \\ & \Rightarrow \triangle ABE \sim \triangle CFB\end{aligned}$$