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Question
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
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Solution
Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.
Let O be the centre of the circle.
We know that the angle formed in the semicircle is 90°.
Since, ABCD is a rectangle, So
`angleADC = angleDCB = angleABC = angleBAD = 90°`
Therefore, AC and BD are diameter of the circle.
We also know that the intersection of any two diameter is the centre of the circle.
Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
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