Q) 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 ∠POQ = 90°.

Ans: Let’s connect Point O with Point R where tangent PQ touches the circle.

Let’s consider ∠ POR = x and ∠ QOR = y

Step 1: Let’s take Δ BOP and Δ ROP:

Here, PB = PR ( tangents from a point on a circle)

OP = OP (common side)

OB = OR (radii of same circle)

∴ Δ BOP Δ ROP

∴ ∠ BOP = ∠ ROP = x ………. (i)

Step 2: Next, we take Δ QOA and Δ QOR:

Here, QA = QR ( tangents from a point on a circle)

OQ = OQ (common side)

OA = OR (radii of same circle)

∴ Δ QOA Δ QOR

∴ ∠ QOA = ∠ QOR = y ……… (ii)

Step 3: Next, We can see from the diagram, that AB is a straight line.

∴ ∠ QOA + ∠ QOR + ∠ ROP + ∠ BOP = 180⁰

∴ y + y + x + x = 180⁰

∴ 2 (x + y) = 180⁰

∴ (x + y) = 90⁰ …. (iii)

Step 4: ∠ POQ = ∠ ROP + ∠ ROQ = x + y

comparing with equation (iii), we get:

∠ POQ = 90⁰

Hence Proved !

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