Question

prove that the parallelogram circumscribing a circle is a rhombus

Answer 1

Radius of the circle is equal everywhere
Now that means if a circle is circumscribed then all the distances from centre to the edge must be equal
So that is diagonals will bisect each other
That is all the 4 parts will be equal.
Diagonals of rhombus are equal and bisect each other.
Hence a parallelogram circumscribing a circle is a rhombus.

Answer 2

Answer:

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DATA :- ABCD is a parallelogram in which its sides are tangents to the circle.

PROVE :- ABCD is a rhombus

PROOF :- we know that tangents drawn from an external point are equal.

•°• AP = AS

BP = BP

CR = CQ

DR = DS

adding LHS & RHS

AP+BP+CR+DR = AS+DS+CQ+AB+DC

= AD+BC

But AB = DC % AD = BC

(opposite sides of parallelogram are equal)

=> AB+AB = BC+BC

2AB = 2BC

AB = BC

similarly AD = DC

=> AB = BC = AD = DC

•°• a parallelogram with all the sides equal is rhombus

=> ABCD a rhombus.