Question

If a quadrilateral ABCD is drawn to
circumscribe a circle, then AB+CD is equal to
(A) AC+BC (B) AD+BC (C) AC+AD
(D) AC+BD+BC​

Answer 1

Given:- Let ABCD be the quadrilateral circumscribing the circle with center 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.

Therefore,

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.

Answer 2

Answer:

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.