if all side of a parallelogram touch a circle,show that the parallelogram is rhombus
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Let ABCD be a parallelogram which circumscribes the circle.
AP = AS [Tangents drawn from an external point to a circle are equal in length]
BP = BQ [Tangents drawn from an external point to a circle are equal in length]
CR = CQ [Tangents drawn from an external point to a circle are equal in length]
DR = DS [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 [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.
AP = AS [Tangents drawn from an external point to a circle are equal in length]
BP = BQ [Tangents drawn from an external point to a circle are equal in length]
CR = CQ [Tangents drawn from an external point to a circle are equal in length]
DR = DS [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 [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.
Answered by
1
The answer for this question is ( the ABCD is a parallelogram.
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