Q28 of 47 Page 1

Prove that a parallelogram circumscribing a circle is a rhombus.


Given: ABCD is a parallelogram.


To prove: ABCD is a rhombus.


Proof: In a parallelogram, opposite sides are equal.


⇒ AB = CD and AD = BC … (1)


We know that tangents drawn from an external point are equal in length.


So, AP = AS


PB = BQ


CR = CQ


DR = DS


By adding these tangents,


⇒ (AP + PB) + (CR + DE) = AS + BQ + CQ + DS


⇒ AB + CD = (AS + DS) + (DQ + CQ)


⇒ AB + CE = AD + BC


From (1),


⇒ AB + AB = BC + BC


⇒ 2AB = 2BC


∴ AB = BC … (2)


From (1) and (2),


AB = BC = CD = AD


∴ Parallelogram ABCD is a rhombus.


Ans. Hence, a parallelogram circumscribing a circle is a rhombus.


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