Question

A circle touches a sides of quadrilateral ABCD prove that AB$+$CD$=$BC$+$DA.

Answer 1

Let the points of contact be $P,Q,R$ and $S$

Two tangents drawn to a circle from the same point are equal in length.

Thus, $AP=AS$

$DS=DR$

$CR=CQ$

$BQ=BP$

$L.H.S.$

$=AB+CD$

$=AP+BP+CR+DR$

$R.H.S.$

$=BC+AD$

$=BQ+CQ+AS+DS$

From the above equalities, $L.H.S.=R.H.S.$