In Figure $\displaystyle 2$, 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 $\displaystyle 2$

\[\begin{aligned} & \triangle O Q P \cong \triangle O R P \Rightarrow \angle Q P O=\angle R P O=45^{\circ} \\ & \Rightarrow \angle Q P R=90^{\circ} . \text { Also } \angle O Q P=\angle O R P=\angle Q O R=90^{\circ} \end{aligned} \] Also \(\displaystyle O R=O Q\). This implies \(\displaystyle O R P Q\) is a square.