Question

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer 1

please refer to the attachment

Answer 2

SOLUTION

Consider a ||gm ABCD circumscribes a circle with centre O.

Consider a ||gm ABCD circumscribes a circle with centre O.We have to prove that AB=BC=CD=AD

We know that length of the tangents drawn from an external points to a circle are equal.

Therefore,

AP= AS

AP= ASBP=BQ

AP= ASBP=BQCR=CQ

AP= ASBP=BQCR=CQDR=DS

Now,

AB+ CD= AD+BC

=) 2AB= 2CD (opposite sides of a ||gm)

=) AB= AD

Hence, CD= AB=AD =BC

=) ABCD is a Rhombus.

hope it helps ☺️