Question

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

Answer 1

We know that the lengths of tangents drawn from external point are equal.

Hence $AP=AS$ ...$\left(1\right)$

$BP=BQ$ ......$\left(2\right)$

$CR=CQ$ ......$\left(3\right)$

$DR=DS$ ......$\left(4\right)$

Adding $\left(1\right)+\left(2\right)+\left(3\right)+\left(4\right)$ we get

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

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

$\Rightarrow AB+CD=AD+BC$

Hence proved.