Question

A circle touches all the four sides of quadrilateral ABCD. Prove that AB + CD = AD + BC.

Answer 1

Given circle touching sides of ABCD at P,Q,R and S 

To prove- 

AB + CD = AD + BC

Proof- 

AP = AS-------(1) tangents from an external point 

PB = BQ-------(2) to a circle are equal in length

DR = DS------(3) 

CR = CQ------(4)

Adding eqn (1),(2),(3) & (4) 

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

AB + DC = AD + BC