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Prove that a parallelogram circumscribing a circle is a rhombus

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Solution:

Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. 

We know that the tangents drawn to a circle from an exterior point are equal in length. 

∴ AP = AS………….(1) 

BP = BQ……………(2) 

CR = CQ …………...(3) 

DR = DS……………(4). 

Adding (1), (2), (3) and (4) 

we get AP+BP+CR+DR = AS+BQ+CQ+DS 

(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) 

∴ AB+CD=AD+BC-----------(5) 

Since AB=DC and AD=BC (opposite sides of parallelogram ABCD) 

putting in (5) we get, 2AB=2AD or AB = AD. 

∴ AB=BC=DC=AD 

Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a rhombus. 

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