Question

If the sides of a quadrilateral ABCD touch a circle, prove that :

AB + CD = BC + AD.

Answer 1

Let AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. Since, tangents to a circle from an exterior point are equal in length, therefore,

AP = AS, BP = BQ, DR = DS and CR = CQ.

On adding, we get :

AP + BP + DR + CR

= AS + BQ + DS + CQ

$\Rightarrow$ AB + CD = AD + BC.