Two circles intersect each other at point 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.

Join AB.

As, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PQ is a tangent and AB is a chord.

∴ ∠QPA = ∠PBA [Angles in alternate segment are equal] ………(1)

Also,

∴ ∠PQA = ∠QBA [Angles in alternate segment are equal] ……….(2)

Adding (1) and (2) we get,

⇒ ∠QPA + ∠PQA = ∠PBA + ∠QBA ……….(3)

⇒ ∠PBA + ∠QBA = ∠PBQ ………..(4)

In △PAQ,

⇒ ∠QPA + ∠PQA + ∠PAQ = 180° [Angle sum property of triangle]

⇒ ∠QPA + ∠PQA = 180° - ∠PAQ

⇒ ∠PBA + ∠QBA = 180° - ∠PAQ [From (3)] ……….(5)

From (4) and (5), we get :

⇒ ∠PBQ = 180° - ∠PAQ

⇒ ∠PBQ + ∠PAQ = 180°.

Hence, proved that PAQ and PBQ are supplementary.