Question

prove that the parallelogram circumscribing a circle is a rhombus.

Answer 1

this is ur answer and it's accurately correct

Answer 2

Hello,

Given:-)A parallelogram ABCD circumscribing a circle with O centre.

TO PROVE:-)AB=BC=CD=AD

Proof:-)We know that the length of tangents drawn from an exterior point to a circle are equal.
so,

AP=AS

BP=BQ

CR=CQ

DR=DS

so,

AB+CD=AP+BP+CR+DR
=AS+BQ+CQ+DS
=(AS+DS)+[BQ+CQ]
Hence,

AB+CD=AD+BC

So,
2AB=2AD

cancel 2 And 2

then,
AB=AD

so,

Now,

CD=AB=AD=BC


Hence,

ABCD Is a rhombus.



that's all

By Sujeet Yaduvanshi.