In the given figure, O is the centre of the circle and QPR is a tangent to it at P . Prove that $\displaystyle \angle \mathrm{QAP}+\angle \mathrm{APR}=90^{\circ}$.

OA = OP \[\therefore \text { In } \triangle \mathrm{OAP}, \angle \mathrm{OPA}=\angle \mathrm{OAP} \] \[\Rightarrow \angle \mathrm{OPA}+\angle \mathrm{APR}=90^{\circ} \] \[\Rightarrow \angle \mathrm{OAP}+\angle \mathrm{APR}=90^{\circ} \] Using (i)