$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F.$ Show that $\Delta ABE \sim \Delta CFB$.
(N/A) In $\triangle ABE$ and $\triangle CFB$:
$1$. $\angle A = \angle C$ (Opposite angles of a parallelogram are equal).
$2$. $\angle AEB = \angle CBF$ (Since $AE \parallel BC$ and $BE$ is a transversal,these are alternate interior angles).
Therefore,by the $AA$ (Angle-Angle) similarity criterion,$\triangle ABE \sim \triangle CFB$.
$1$. $\angle A = \angle C$ (Opposite angles of a parallelogram are equal).
$2$. $\angle AEB = \angle CBF$ (Since $AE \parallel BC$ and $BE$ is a transversal,these are alternate interior angles).
Therefore,by the $AA$ (Angle-Angle) similarity criterion,$\triangle ABE \sim \triangle CFB$.
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