Prove that the circles described with the four sides of a rhombus, as diameters,

Question:

Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.

Solution:

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O.

The diagonals of a rhombus bisect each other at right angles.
i.e., ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
Now, circles with AB, BC, CD and DA as diameter passes through O (angle in a semi-circle is a right angle).
Hence, the circle with four sides of a rhombus as diameter, pass through O, i.e., the point of intersection of its diagonals.

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Prashant Kumar
Feb. 19, 2024, 6:35 a.m.
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Uuhzz
Jan. 24, 2024, 6:35 a.m.
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Shubham kumar
Feb. 2, 2024, 8:08 a.m.
Gv
Bhavana Gupta
Jan. 21, 2024, 6:35 a.m.
Didn't understand anything
Bhavana Gupta
Jan. 21, 2024, 6:35 a.m.
Didn't understand anything
Urvi
Jan. 13, 2024, 6:35 a.m.
Good answer as compared to others
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