Question

Question 7
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer 1

Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.

We know that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\angle BAD=\frac{1}{2}\angle BOD=\frac{180}{2}={90}^{\circ }$
$\angle BCD+\angle BAD={180}^{\circ }$ (Opposite angles of a cyclic quadrilateral)
$\angle BCD={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

We know that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\angle ADC=\frac{1}{2}\angle AOC=\frac{1}{2}\left({180}^{\circ }\right)={90}^{\circ }$
$\angle ADC+\angle ABC={180}^{\circ }$ (Opposite angles of a cyclic quadrilateral)
${90}^{\circ }+\angle ABC={180}^{\circ }$
$\angle ABC={90}^{\circ }$
Each interior angle of a cyclic quadrilateral is of ${90}^{\circ }$. Hence, it is a rectangle.