Prove that the parallelogram circumscribing a circle is a rhombus?this question is from the lesson Circles of 10th std

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The tangency points of ABCD parallelogram are PQMN.

P `in`  AB ; Q `in`  BC ; M`in`  CD ; N `in`  AD.

To prove that the parallelogram circumscribing the circle is a rhombus, you need to prove that all side of parallelogram are equal.

Let's take the triangles OPA and ONA.

The sides OP and ON are perpendicular to AB, respectively to AD (the circle radii are perpendicular to the tangent lines to circle).

OP`_|_` AB ; ON `_|_`  AD

OP `-=` ON (radii of the circle)

AO is the common side of both triangles.

In accordance with all the above, the triangles are equal => AP`-=` AN

Use the same approach to prove BP`-=` BQ; CQ`-=` QM ; DM`-=` DN.

Use the fact that diagonals of a parallelogram bisect each other to prove that AP`-=` PB; BQ`-=` QC ; CM`-=` MD ; DN`-=` AN.

But AP`-=` AN and DN`-=` AN => AB`-=` BC`-=` CD`-=` AD

The parallelogram with equal sides is a rhombus, therefore ABCD is a rhombus.

ANSWER: The parallelogram ABCD that circumscribes the circle is a rhombus.