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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 $∆ABE \sim ∆CFB$.
Given:
$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$.
To do:
We have to show that $∆ABE \sim ∆CFB$.
Solution:
In the above figure, $ABCD$ is a parallelogram in which $E$ is a point on $AD$ produced and $BE$ intersects $CD$ at $F$.
In parallelogram $ABCD$,
$\angle A=\angle C$.......(i) (opposite angles)
In $\triangle ABE$ and $\triangle CFB$,
$\angle EAB=\angle BCF$ (Alternate angles)
$\angle ABE=\angle BFC$
Therefore, by AA criterion,
$\triangle ABE \sim \triangle CFB$
Hence proved.
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