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

Solution: Please refer to diagram in the video.
As we know, tangents drawn from a point to a circle are always equal.So,
$CR=CQ,PB=BQ,AP=AS\text{and}DR=DS$
So,$CR+PB+AP+DR=CQ+BQ+AS+DS$
$\left(CR+DR\right)+\left(PB+AP\right)=\left(CQ+BQ\right)+\left(AS+DS\right)$
As, $CS+DR=CD$
$PB+AP=AB$
$CQ+BQ=BC$
$AS+DS=AD$
So,$AB+CD=AD+BC$