In the figure, the sides $BA$ and $CA$ have been produced such that $BA = AD$ and $CA = AE$. Prove the segment $DE \parallel BC$.
"
Given:
The sides $BA$ and $CA$ have been produced such that $BA = AD$ and $CA = AE$.
To do:
We have to prove the segment $DE \parallel BC$.
Solution:
In $\triangle ABC$ and $\triangle DAE$,
$AB=AD$ (Given)
$AC = AE$ (Given)
$\angle BAC = \angle DAE$ (Vertically opposite angles)
Therefore, by SAS axiom, we get,
$\triangle ABC \cong \triangle DAE$
This implies,
$\angle ABC = \angle ADE$ (CPCT)
$\angle ABC$ and $\angle ADE$ are alternate angles.
Therefore,
$DE \parallel BC$.
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