Question

In the figure given below a quadrilateral ABCD is draw 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:-

$AP=AS.....\left(1\right)(\because \text{Length of tangents drawn from external point are equal})$

Similarly,

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

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

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

Adding equation $\left(1\right),\left(2\right),\left(3\right)\&\left(4\right)$, we have

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

$(AP+PB)+(CR+RD)=(AS+SD)+(BQ+QC)$

$AB+CD=AD+BC$

Hence proved.