Let us consider 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 Proof : Sum of opposite sides are equal, i.e. 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.