Question

Two circles intersect each other at points A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.

Answer 1

Join AB.

PQ is the tangent, and AB is a chord

∴ ∠QPA = ∠PBA  ...(i) (Angles in alternate segment)

Similarly,

∠PQA = ∠QBA  ...(ii)

Adding (i) and (ii)

∠QPA + ∠PQA = ∠PBA + ∠QBA

But, in ΔPAQ,

∠QPA + ∠PQA = 180° – ∠PAQ  ...(iii)

And ∠PBA + ∠QBA = ∠PBQ  ...(iv)

From (iii) and (iv)

∠PBQ = 180° – ∠PAQ

⇒ ∠PBQ + ∠PAQ = 180°

⇒ ∠PBQ + ∠PBQ = 180°

Hence, ∠PAQ and ∠PBQ are supplementary.