Prove that the parallelogram circumscribing a circle is a rhombus.

Given: A paralalogram circumscribing a circle.

To do:To prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

$\because$ 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.

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

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

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

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

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

$AP+BP+CR+DR=AS+BQ+CQ+DS$

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

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

 

$\Rightarrow AB+CD=AD+BC$ 

$\Rightarrow AB+AB=BC+BC\ $ $( \because \ ABCD\ is\ a\ parallelogram. AB=CD\ and\ BC=AD\ )$

$\Rightarrow 2AB=2BC$

Thus $AB=BC=CD=AD$

It proves that the given parallelogram is a rhombus.