Answer :

Let ABCD be a rhombus in which diagonals intersect at point O and a circle is drawn is drawn taking side CD as diameter

Now,

∠COD = 90^{0}

(Since, diagonals of rhombus intersect at 90^{0})

Hence,

∠AOB = ∠BOC = ∠COD = ∠DOA = 90^{0}

Hence, point O has to lie on the circle

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