Given: A quadrilateral ABCD, And a circle is circumscribed by ABCD

Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.

To Prove: AB + CD = AD + BC

Proof:

In the Figure,

As tangents drawn from an external point are equal.

We have

AP = AS [tangents from point A]

BP = BQ [tangents from point B]

CR = CQ [tangents from point C]

DR = DS [tangents from point D]

Add the above equations

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

⇒ AB + CD = AS + DS + BQ + CQ

⇒ AB + CD = AD + BC

Hence Proved.