102
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 $\angle POQ = 90^\circ$.
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Join OR.
△ AOQ ≅ △ ROQ ⇒ ∠ AOQ = ∠ ROQ$ ----- (i) $△ BOP ≅ △ ROP ⇒ ∠ BOP = ∠ ROP$ ----- (ii) Since $∠ AOR + ∠ ROB = 180^°$ $⇒ 2∠ QOR + 2∠ ROP = 180^°$ $⇒ ∠ QOR + ∠ ROP = ∠ POQ = 90^°_a_1.png)
△ AOQ ≅ △ ROQ ⇒ ∠ AOQ = ∠ ROQ$ ----- (i) $△ BOP ≅ △ ROP ⇒ ∠ BOP = ∠ ROP$ ----- (ii) Since $∠ AOR + ∠ ROB = 180^°$ $⇒ 2∠ QOR + 2∠ ROP = 180^°$ $⇒ ∠ QOR + ∠ ROP = ∠ POQ = 90^°