Prove that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Explanation:
Given, \(ABCD\) is a rectangle inscribed in a circle.
Let the diagonals \(AC\) and \(BD\) intersect at the point \(O\).
\(OA=\) and \(OB =\) []
This implies, \(O\) is the .
Also, we know that, every angle in a rectangle is \(^\circ\).
Since, the rectangle is inscribed in a circle, the diagonals \(AC\) and \(BD\) are the of the circle.
So, the chords \(AC\) and \(BD\) subtends a at the circumference.
Thus, the diagonals \(AC\) and \(BD\) are the of the circle. [Since, ]
Therefore, the point \(O\) is the of the circle.
Therefore, the diagonals of the rectangle \(ABCD\) intersect at the centre.