Question

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB = CD =
AD = BC​

Answer 1

Answer:

Given:-  Let ABCD be the quadrilateral circumscribing the circle with centre O.          The quadrilateral touches the circle at point P,Q,R and S.

To prove:-  AB+CD+AD+BC

Proof:-

As we know that, length of tangents drawn from the external point are equal.

∴ , AP=AS.....(1)

BP=BQ.....(2)

CR=CQ.....(3)

DR=DS.....(4)

Adding equation (1),(2),(3) and (4), we get

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

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

AB+CD=AD+BC

Hence proved.