Question

prove that the circle drawn with any side of rhombus as diameter passes through the point of intersection of its diagonal​

Answer 1

Answer: Consider ABCD be a rhombus whose diagonals AC and BD intersect at O.

Since diagonals of a rhombus intersect each other at right angle.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, AB, BC, CD and DA as diameters of circles passes through O. Angle in a semi-circle is 90°.

Hence the circles described on the four sides of a rhombus as diameter, passes through the point of intersection of its diagonals.

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Step-by-step explanation:

Answer 2

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