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