Question

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 $O$ which is centre of the circle.

Consider, $\Delta ABC$,as $AC$ is the diameter, therefore,

$\Rightarrow \angle ABC={90}^{\circ }$ (Angle in a semi circle is ${90}^{\circ }$)

Consider, $\Delta BCD$,as $DB$ is the diameter, therefore,

$\Rightarrow \angle DCB={90}^{\circ }$ (Angle in a semi circle is ${90}^{\circ }$)

Since, $ABCD$ be a cyclic quadrilateral, therefore,

$\Rightarrow \angle ADC={180}^{\circ }-\angle ABC={90}^{\circ }$ and

$\Rightarrow \angle DAB={180}^{\circ }-\angle DBC={90}^{\circ }$

Since, all the angles of cyclic quadrilateral $ABCD$ are ${90}^{\circ }$, therefore,

$ABCD$ is a rectangle.