Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

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#### Solution

Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.

∴ ∠COD = 90°

Also, in rhombus, the diagonals intersect each other at 90°.

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Clearly, point O has to lie on the circle.

Concept: Cyclic Quadrilateral

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