Let us prove that if the circles are drawn having sides of a rhombus as diameter then the circles pass through the fixed points.
( Please show with diagram )
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Step-by-step explanation:
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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