In the given figure,$AB = AC$ and $BP = CQ$. Prove that $\Delta APQ$ is an isosceles triangle.
(N/A) In $\triangle ABC$,$AB = AC$ (Given).
$\therefore \angle ABC = \angle ACB$ ($\because$ Angles opposite to equal sides are equal).
Since $P$ lies on $BC$ and $Q$ lies on $BC$,we have $\angle ABP = \angle ACQ$.
Now,in $\triangle ABP$ and $\triangle ACQ$:
$AB = AC$ (Given)
$BP = CQ$ (Given)
$\angle ABP = \angle ACQ$ (Proved above)
So,by $SAS$ congruence criterion,$\triangle ABP \cong \triangle ACQ$.
$\therefore AP = AQ$ (by $CPCT$).
Now,in $\triangle APQ$,since $AP = AQ$,$\Delta APQ$ is an isosceles triangle.
$\therefore \angle ABC = \angle ACB$ ($\because$ Angles opposite to equal sides are equal).
Since $P$ lies on $BC$ and $Q$ lies on $BC$,we have $\angle ABP = \angle ACQ$.
Now,in $\triangle ABP$ and $\triangle ACQ$:
$AB = AC$ (Given)
$BP = CQ$ (Given)
$\angle ABP = \angle ACQ$ (Proved above)
So,by $SAS$ congruence criterion,$\triangle ABP \cong \triangle ACQ$.
$\therefore AP = AQ$ (by $CPCT$).
Now,in $\triangle APQ$,since $AP = AQ$,$\Delta APQ$ is an isosceles triangle.
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