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 : AB + CD = AD + BC
In the Figure,
As tangents drawn from an external point are equal.
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
Rate this question :