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

Prove that the parallelogram circumscribing a circle is a rhombus.

 

 

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To prove - Parallelogram circumscribing circle is a rhombus.

We need to frame ABCP is a rhombus.

Theoram - lengths of tangents drawn from external point are equal Hence, using this theorem.

EP=HP ____(ii)

AE=AF _____(iii)

BG=BF _____(iv)

GC=CH ______(v)

(ii), (iii), (iv), (v)

EP+AE+BG+GC=HP+AF+BF+CH

AP+BC=AB+PC

 AP=BC & AB=CP (parallelogram) 

2BC=2CP

BC= CP

Similarly AP=AB=BC=CP

Hence all sides are equal ABCP is a rhombus.

Hence proved

Posted by

Safeer PP

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