92
In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that $\angle POQ = 90^\circ$.
_q_1.png)
Show SolutionHide Solution↓
Sol. Join OR.
$\triangle AOQ \cong \triangle ROQ \Rightarrow \angle AOQ = \angle ROQ$ ----- (i)
$\triangle BOP \cong \triangle ROP \Rightarrow \angle BOP = \angle ROP$ ----- (ii)
Since $\angle AOR + \angle ROB = 180^\circ$
$\Rightarrow 2\angle QOR + 2\angle ROP = 180^\circ$
$\Rightarrow \angle QOR + \angle ROP = \angle POQ = 90^\circ$_a_1.png)
$\triangle AOQ \cong \triangle ROQ \Rightarrow \angle AOQ = \angle ROQ$ ----- (i)
$\triangle BOP \cong \triangle ROP \Rightarrow \angle BOP = \angle ROP$ ----- (ii)
Since $\angle AOR + \angle ROB = 180^\circ$
$\Rightarrow 2\angle QOR + 2\angle ROP = 180^\circ$
$\Rightarrow \angle QOR + \angle ROP = \angle POQ = 90^\circ$