Question

prove that a parallelogram circumscribed by a circle is a rhombus.

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 + BC Hence 2AB = 2BC Therefore, AB = BC Similarly, we get AB = DA and DA = CD Thus ABCD is a rhombus.

Answer 2

Answer:

Step-by-step explanation:

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