In the given figure,$XP = XS$,$XQ = XR$ and $\angle PXR = \angle SXQ$. Prove that $PQ = SR$.

(N/A) Given: $\angle PXR = \angle SXQ$
Subtracting $\angle QXR$ from both sides:
$\angle PXR - \angle QXR = \angle SXQ - \angle QXR$
$\therefore \angle PXQ = \angle SXR \quad \dots(1)$
Now,in $\Delta XPQ$ and $\Delta XSR$:
$XP = XS$ (Given)
$XQ = XR$ (Given)
$\angle PXQ = \angle SXR$ [From $(1)$]
$\therefore \Delta XPQ \cong \Delta XSR$ (By $SAS$ congruence criterion)
$\therefore PQ = SR$ (By $CPCT$)