In the figure, $ABCD$ is a parallelogram and $E$ is the mid-point of side $BC$. If $DE$ and $AB$ when produced meet at $F$, prove that $AF = 2AB$.
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Given:

$ABCD$ is a parallelogram and $E$ is the mid-point of side $BC$. 

$DE$ and $AB$ when produced meet at $F$.

To do:

We have to prove that $AF = 2AB$.

Solution:

In $\triangle CDE$ and $\triangle EBF$,

$\angle DEC = \angle BEF$                (Vertically opposite angles)

$CE = EB$                                               ($E$ is the mid point of $BC$)

$\angle DCE = \angle EBF$                (Alternate angles)

Therefore, by SAS axiom,

$\triangle CDE \cong \triangle EBF$

This implies,

$DC = BF$               (CPCT)

$AB = DC$                    (Opposite sides of a parallelogram)

Therefore,

$AB = BF$

$AF = AB + BF$

$= AB + AB$

$= 2AB$

Hence, $AF = 2AB$.

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Updated on: 10-Oct-2022

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