Question

Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that : ∠PAQ = 2∠OPQ

Answer 1

In quadrilateral OPAQ,

∠OPA = ∠OQA = 90°

(∵ OP ⊥ PA and OQ ⊥ QA)

∴ `∠`POQ + `∠`PAQ + 90° + 90° = 360°

`=>` ∠POQ + ∠PAQ = 360° – 180° = 180°   ...(i)

In triangle OPQ,

OP = OQ  ...(Radii of the same circle)

∴ OPQ = ∠OQP

But ∠POQ + ∠OPQ + ∠OQP = 180°

`=>` ∠POQ + ∠OPQ + ∠OPQ = 180°

`=>` ∠POQ + 2∠OPQ = 180°   ...(ii)

From (i) and (ii)

∠POQ + ∠PAQ = ∠POQ + 2∠OPQ

`=>` ∠PAQ = 2∠OPQ