Question

A quadrilateral ABCD is drawn to circumscribe a circle. (see the figure) 2

Prove that AB + CD = AD + BC​

Answer 1

Answer:

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.

Therefore,

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:

..quad.ABCD AB //CD

..AC=BD

..

...ABCD ISA QUAD.SO

AC//BD

..AB=CD

... AB+CD=AD+BC