Sum

Prove that a parallelogram circumscribing a circle is a rhombus.

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#### Solution

**Given:** ABCD be a parallelogram circumscribing a circle with centre O.

**To prove:** ABCD is a rhombus.

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

∴ AP = AS, BP = BQ, CR = CQ and DR = DS.

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

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

∴ AB + CD = AD + BC or 2AB = 2BC (since AB = DC and AD = BC)

∴ AB = BC = DC = AD.

Therefore, ABCD is a rhombus.

**Hence, proved.**

Concept: Number of Tangents from a Point on a Circle

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