Question

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 )​

Answer 1

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$\huge \purple {ANSWER}$

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Let ABCD be a rhombus whose diagonals AC and BD intersect at O.

We know that, diagonals of a rhombus intersect each other at right angle.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, circles with AB, BC, CD and DA as their respective diameters pass through O. (Angle in a semi-circle is 90°)

Thus, the circles described on the four sides of a rhombus as their respective diameters, pass through the point of intersection of its diagonals.

Hence, proved

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Answer 2

Answer:

Nahi sorry, maine galti se likh diya tha ....!!