Question

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer 1

the solution is given below
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




hope u understood

Answer 2

Answer:

Step-by-step explanation:

the solution is given below

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