Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.

The two intersecting circles with a secant passing through P intersecting the circles in A and B are shown in the figure below:

We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in alternate segment.

∴ ∠TAP = ∠AQP (Angles in alternate segment are equal) ………(1)

and, ∠TBP = ∠BQP …………….(2)

Adding (1) and (2), we get :

⇒ ∠TAP + ∠TBP = ∠AQP + ∠BQP

⇒ ∠TAP + ∠TBP = ∠AQB …………..(3)

In △TAB,

⇒ ∠ATB + ∠TAB + ∠TBA = 180° [By angle sum property of triangle]

⇒ ∠ATB + ∠TAP + ∠TBP = 180° [From figure, ∠TAB = ∠TAP and ∠TBA = ∠TBP]

⇒ ∠ATB + ∠AQB = 180°. [From (3)]

Since, sum of opposite angles = 180°,

which is possible when the quadrilateral is cyclic quadrilateral.

Hence, proved that A, Q, B and T lie on a circle.