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Prove that the centre of the circle circumscribing the cyclic rectangle $ABCD$ is the point of intersection of its diagonals.
To do:
We have to prove that the centre of the circle circumscribing the cyclic rectangle $ABCD$ is the point of intersection of its diagonals.
Solution:
$ABCD$ is a cyclic rectangle and diagonals $AC$ and $BD$ intersect each other at $O$.
Let $O$ be the centre of the circle circumscribing the rectangle $ABCD$.
Each angle of a rectangle is a right angle and $AC$ is the chord of the circle.
Therefore,
$AC$ is the diameter of the circle.
Similarly, we can prove that diagonal $BD$ is also the diameter of the circle.
This implies,
The diameters of the circle pass through the centre.
Hence the point of intersection of the diagonals of the rectangle is the centre of the circle.
Updated on: 10-Oct-2022
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