In the figure,$D$ and $E$ are points on side $BC$ of a $\triangle ABC$ such that $BD = CE$ and $AD = AE$. Show that $\triangle ABD \cong \triangle ACE$.

(N/A) Given: $\triangle ABC$ in which $BD = CE$ and $AD = AE$.
To Prove: $\triangle ABD \cong \triangle ACE$.
Proof: In $\triangle ADE$,we have
$AD = AE$ [Given]
$\Rightarrow \angle ADE = \angle AED$ [Angles opposite to equal sides of a triangle are equal]
Now,$\angle ADB + \angle ADE = 180^{\circ}$ [Linear pair axiom]
$\angle AEC + \angle AED = 180^{\circ}$ [Linear pair axiom]
From these equations,we get:
$\angle ADB + \angle ADE = \angle AEC + \angle AED$
Since $\angle ADE = \angle AED$,we have $\angle ADB = \angle AEC$.
Now,in $\triangle ABD$ and $\triangle ACE$:
$AD = AE$ [Given]
$\angle ADB = \angle AEC$ [Proved above]
$BD = CE$ [Given]
So,by $SAS$ criterion of congruence,we have
$\triangle ABD \cong \triangle ACE$.
Hence,proved.