In the figure, $AE$ bisects $\angle CAD$ and $\angle B = \angle C$. Prove that $AE \parallel BC$.
"
Given:
$AE$ bisects $\angle CAD$ and $\angle B = \angle C$.
To do:
We have to prove that $AE \parallel BC$.
Solution:
In $\triangle ABC, BA$ is produced.
This implies,
$\angle CAD = \angle B + \angle C$
$2\angle EAC = \angle C + \angle C$ (Since $AE$ is the bisector of $\angle CAE$)
$2\angle EAC = 2\angle C$
$\angle EAC = \angle C$
$\angle EAC$ and $\angle C$ are alternate angles
This implies,
$AE \parallel BC$.
Hence proved.
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