In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with center O, in such a way that the sides AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. Prove That : AB + CD = BC + DA.

Given : A quadrilateral ABCD is drawn to circumscribe a circle with center O, in such a way that the sides AB, BC, CD and DA touch the circle at points P, Q, R and S respectively.

To prove : AB + CD = BC + DA

Proof :

PB and BQ are tangents of circle passing through points P and Q respectively and are from an external point B.

CR and QC are tangents of circle passing through points Q and R respectively and are from an external point C.

RD and DS are tangents of circle passing through points R and S respectively and are from an external point D.

AP and SA are tangents of circle passing through points S and P respectively and are from an external point A.

From all these points we get

PB = BQ [eqn 1]

CR = QC [eqn 2]

RD = DS [eqn 3]

AP = SA [eqn 4]

[As, Tangents drawn from an external point to a circle are equal. ]

Add eqn 1, eqn 2, eqn 3 and eqn 4

PB + CR + RD + AP = BQ + QC + DS + SA

AP + PB + CR + RD = BQ + QC + DS + SA

AB + CD = BC + DA

Hence proved.