Question

In the figure (i) given below, the sides of the quadrilateral touch the circle Prove that $AB+CD=BC+DA$

Answer 1

Lengths of the tangents from an external point are equal.

$AP=AS$, $BP=BQ$, $CR=CQ$ and $DR=DS$

$AB+CD=(AP+BP)+(CR+DR)$

$=AS+BQ+CQ+DS$

$=(AS+DS)+(BQ+CQ)$

$=AD+BC$

Hence $AB+CD=AD+BC$.