Question

A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig). prove that $AB+CD=AD+BC$.

Answer 1

It is known that the length of the tangents drawn from the external point are always equal, so,

$PB=QB$

$QC=RC$

$AP=AS$

$DS=DR$

Now,

$AB+CD$

$=AP+PB+DR+RC$

$=AS+QB+DS+CQ$

$=AS+DS+QB+CQ$

$=AD+BC$