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 $\angle POQ = 90°$.

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Question

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 $\angle POQ = 90°$.

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Accepted Answer

Step 1: Join OR.
$	riangle AOQ \cong 	riangle ROQ \Rightarrow \angle AOQ = \angle ROQ$ ... (i)
Step 2: $	riangle BOP \cong 	riangle ROP \Rightarrow \angle BOP = \angle ROP$ ... (ii)
Step 3: Since $\angle AOR + \angle ROB = 180°$
$\Rightarrow 2\angle QOR + 2\angle ROP = 180°$
$\Rightarrow \angle QOR + \angle ROP = 90°$
$\Rightarrow \angle POQ = 90°$

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