Question

prove that the parallelorgram circumscribing a circle is rhombus​

Answer 1

Proof:

Given :- circle with Centre O and also given ABCD is the llgm

We need to prove that ABCD is a rhombus

pls mark the pt of contact For AB - P BC- Q CD-R DA-S Proof:- Tangents from an external point to a circle are equal in length.

Therfore AP=AS

BP=BQ

DR=DS

CR=CQ

Adding all the LHS and RHS we get

AB+CD=AD+BC eq.1

Since ABCD is llgm

Opp sides will be equal. Therefore AB=CD & AD=BC

when u substitute the above in eq.1 we get 2AB=2AD

=> AB=AD

=> AB=BC=CD=DA

Therefore ABCD is a rhombus

Answer 2

this is one of the basic Derivations taught in Maths in High School. so find a book to understand it better. this is not the right place to understand it well.

any other doubts do not hesitate to ask me directly.