In $\Delta PQR$,$PQ = PR$. Prove that $\angle R = \angle Q$.
(N/A) Draw the bisector $PM$ of $\angle P$ which meets $QR$ at $M$.
Therefore,$\angle QPM = \angle RPM$ $(1)$.
Now,in $\Delta PMQ$ and $\Delta PMR$:
$PQ = PR$ (Given)
$PM = PM$ (Common side)
$\angle QPM = \angle RPM$ [From $(1)$]
So,by $SAS$ congruence criterion,$\Delta PMQ \cong \Delta PMR$.
Therefore,$\angle R = \angle Q$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
Therefore,$\angle QPM = \angle RPM$ $(1)$.
Now,in $\Delta PMQ$ and $\Delta PMR$:
$PQ = PR$ (Given)
$PM = PM$ (Common side)
$\angle QPM = \angle RPM$ [From $(1)$]
So,by $SAS$ congruence criterion,$\Delta PMQ \cong \Delta PMR$.
Therefore,$\angle R = \angle Q$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
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