Question

Prove that a parallelogram circumscribing a circle is a rhombus.
OR
In Figure 2, AD ⊥ BC. Prove that AB² + CD² = BD² + AC² .

Answer 1

Let ABCD be a parallelogram which circumscribes the circle. AP = AS [Since tangents drawn from an external point to a circle are equal in length] BP = BQ [Since tangents drawn from an external point to a circle are equal in length] CR = CQ [Since tangents drawn from an external point to a circle are equal in length] DR = DS [Since tangents drawn from an external point to a circle are equal in length] Consider, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)AB + CD = AD + BC But AB = CD and BC = AD  [Since opposite sides of parallelogram ABCD] AB + CD = AD + BCHence 2AB = 2BC Therefore, AB = BC  Similarly, we get AB = DA and  DA = CDThus ABCD is a rhombus.

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