A circle touches all the four sides of a quadrilateral $ABCD$. Prove that $AB+CD=BC+DA$.

Given: A circle touches all the four sides of a quadrilateral ABCD.

To do: To prove that $AB+CD=BC+DA$.

$\because $ Given parallelogram $ABCD$ is circumscribing the circle and its sides are touching the circle at P, Q, R and S.

$\therefore $ AP and AS are tangent to the circle from the external point A.

BP and BQ are tangent to to the circle from the external point B.

CQ and CR are tangents to the circle from the external point C.

DR and DS are tangents to the circle from the external point D.

As we know that tangents drawn to a circle from an external point are always of equal length.

$\therefore  AP=AS\ \dotsc \dotsc \dotsc \dotsc .( 1)$

$BP=BQ\dotsc \dotsc \dotsc \dotsc \dotsc \dotsc ( 2)$

$CQ=CR\dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc .( 3)$

$DR=DS\dotsc \dotsc \dotsc \dotsc \dotsc \dotsc .\dotsc ( 4)$

Let us add $( 1) ,\ ( 2) ,\ ( 3)$ and $( 4)$

$AP+BP+CR+DR=AS+BQ+CQ+DS .......... ( \because AP+BP=AB,\ CR+DR=CD,\ AS+DS=AD\ and\ BQ+CQ=BC)$

$\Rightarrow ( AP+BP) +( CR+DR) =( AS+DS) +( BQ+CQ)$ 

$\Rightarrow AB+CD=AD+BC$