Question

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

Answer 1

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.

Therefore,

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

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

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

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

Adding equation $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$, 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.