Question

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

Answer 1

let the circle touches the sides of the quad ABCD at the points P Q R and S

since the length of tangents drawn from an external point to a circle are equal

So AP =AS ,BP=BQ, CR=CQ, nd DR=DS

on addng these

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

ÀB+CD=(AS+DS)+(BQ+CQ)

AB + CD=BC+AD

hope this will help you

Answer 2

Answer:

let the circle touches the sides of the quad ABCD at the points P Q R and S

since the length of tangents drawn from an external point to a circle are equal

So AP =AS ,BP=BQ, CR=CQ, nd DR=DS

on addng these

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

ÀB+CD=(AS+DS)+(BQ+CQ)

AB + CD=BC+AD