In the given figure, the parallelogram ABCD circumscribes a circle, touching the circle at P, Q, R and S. Prove that :

Join AC and BD.

(a) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

∴ AP = AS, BP = BQ, CR = CQ and DR = DS

Adding the equations, we get :

⇒ AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC ......(1)

We know that,

Opposite sides of a parallelogram are equal.

∴ AB = CD and AD = BC

Substituting the above values in equation (1), we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, proved that AB = BC.

(b) From part (a),

⇒ AB = BC

Also, AB = CD and AD = BC [Opposite sides of a parallelogram are equal]

∴ AB = BC = CD = AD

Since, all the four sides of the parallelogram ABCD are equal, ABCD is a rhombus.

We know that,

The diagonals of a rhombus bisect each other at right angles.

Hence, proved that the diagonals AC and BD bisect each other at right angles.