If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Let a circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.

We know that,

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

As, AP and AS are tangents to the circle from an external point A, we have

AP = AS ……… (1)

Similarly, we also get

BP = BQ ……… (2)

CR = CQ ……… (3)

DR = DS ……… (4)

Adding (1), (2), (3) and (4), we get

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

From figure,

AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC

⇒ AB + CD = AD + BC

Hence, proved that AB + CD = AD + BC.