In the figure,$AC = AE$,$AB = AD$ and $\angle BAD = \angle EAC$. Show that $BC = DE$.
(N/A) We have $\angle BAD = \angle EAC$.
Adding $\angle DAC$ on both sides,we have:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$
Now,in $\triangle ABC$ and $\triangle ADE$,we have:
$\angle BAC = \angle DAE$ [Proved above]
$AB = AD$ [Given]
$AC = AE$ [Given]
$\therefore \triangle ABC \cong \triangle ADE$ [Using $SAS$ congruence criterion]
Since $\triangle ABC \cong \triangle ADE$,therefore,their corresponding parts are equal $(CPCT)$.
$\Rightarrow BC = DE$.
Adding $\angle DAC$ on both sides,we have:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$
Now,in $\triangle ABC$ and $\triangle ADE$,we have:
$\angle BAC = \angle DAE$ [Proved above]
$AB = AD$ [Given]
$AC = AE$ [Given]
$\therefore \triangle ABC \cong \triangle ADE$ [Using $SAS$ congruence criterion]
Since $\triangle ABC \cong \triangle ADE$,therefore,their corresponding parts are equal $(CPCT)$.
$\Rightarrow BC = DE$.
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