Question

[rei1rc ail the sides of a parallelogram touch a circle, then the parallelogram is​

Answer 1

Answer:

then the parallelogram is cyclic quadrilateral.

branlist mar kar do bhai please

Answer 2

Answer:

Given : Sides AB,BC,CD andDA of a parrallelogram ABCD touch a circle at P,Q,Rand S respectively.

To prove : Parallelogram ABCD is a rhombus.

Proof : 

SinceTangents drawn from an external point to a circle are equal

AP=AS.....(1)(Tangent from point A )

BP=BQ....(2)(Tangent from point B )

CR=CQ....(3)(Tangent from point A )

DR=DS....(4)(Tangent from point A )

Adding (1), (2), (3) and (4), we get

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

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

AB+CD=AD+BC

AB+AB=AD+AD [In a Parallelogram ABCD, opposite sides are equal]

2AB=2ADorAB=AD

But AB=CDandAD=BC [Opposite sides of a Parallelogram are equal]

AB=BC=CD=DA

Hence, Parallelogram ABCD is a rhombus.

Step-by-step explanation:

plzz mark as brainlist and follow me