$ABCD$ is a cyclic quadrilateral in which $BA$ and $CD$ when produced meet in $E$ and $EA = ED$. Prove that $AD \| BC$.

Given:

$ABCD$ is a cyclic quadrilateral in which $BA$ and $CD$ when produced meet in $E$ and $EA = ED$.

To do:

We have to prove that $AD \| BC$.

Solution:

$EA = ED$

This implies,

$\angle EAD = \angle EDA$               (Angles opposite to equal sides are equal)

In cyclic quadrilateral $ABCD$,

$\angle EAD = \angle C$

Similarly,

$\angle EDA = \angle B$

$EAD = \angle EDA$

Therefore,

$\angle B = \angle C$

In $\triangle EBC$,

$\angle B = \angle C$

This implies,

$EC = EB$            (Sides opposite to equal angles are equal)

$\angle EAD = \angle B$

$\angle EAD$ and $\angle B$ are corresponding angles

Therefore,

$AD \| BC$.

Hence proved .

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Updated on: 10-Oct-2022

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