In Figure $\displaystyle 1$, PQ and PR are tangents to the circle centred at O. If $\displaystyle \angle \mathrm{OPR}=45^{\circ}$, then prove that ORPQ is a square. Figure $\displaystyle 1$

\[\triangle O Q P \cong \triangle O R P \Rightarrow \angle Q P O=\angle R P O=45^{\circ} \] $\displaystyle \Rightarrow \angle Q P R=90^{\circ}$. Also $\displaystyle \angle O Q P=\angle O R P=\angle Q O R=90^{\circ}$ Also $\displaystyle O R=O Q$. This implies $\displaystyle O R P Q$ is a square.