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11.  Prove that the parallelogram circumscribing a circle is a rhombus.

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To prove -  the parallelogram circumscribing a circle is a rhombus
Proof-
ABCD is a parallelogram which circumscribes a circle with centre O.
P, Q, R, S are the points of contacts on sides AB, BC, CD and DA  respectively
AB = CD .and AD = BC...........(i)
It is known that, tangents drawn from an external points are equal in length.
RD = DS ...........(ii)
RC = QC...........(iii)
BP = BQ...........(iv)
AP = AS .............(v)
By adding eq (ii) to eq (v) we get;
(RD + RC) + (BP + AP) = (DS + AS) + (BQ + QC)
CD + AB = AD + BC
\Rightarrow 2AB = 2AD [from equation (i)]
\RightarrowAB = AD
Now, AB = AD and AB  = CD 
\therefore AB = AD = CD = BC

Hence ABCD is a rhombus. 

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