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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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Correct figure
In $\triangle ABE$ and $\triangle CFB$
$\angle EAB = \angle BCF$
$\angle AEB = \angle CBF$
$\Rightarrow \triangle ABE \sim \triangle CFB$._a_1.png)
In $\triangle ABE$ and $\triangle CFB$
$\angle EAB = \angle BCF$
$\angle AEB = \angle CBF$
$\Rightarrow \triangle ABE \sim \triangle CFB$