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

Hint:
We know that a rectangle is a parallelogram with one angle ${90^\circ }$. So, we first prove $ABCD$ is a parallelogram and then one angle ${90^\circ }$.

Complete step by step solution:
We have given that $ABCD$ be a cyclic quadrilateral having diagonals $AC$ and $BD$, intersecting each other at point $O$.

We have to prove that $ABCD$ is a rectangle.
We know that a rectangle is a parallelogram with one angle ${90^\circ }$. So, we first prove $ABCD$ is a parallelogram and then one angle ${90^\circ }$.
Since $BD$ is the diameter and hence the arc$BAD$ is in a semi circle.
$\therefore \angle BAD = {90^\circ }$ ….. (1) (Angle in a semicircle is ${90^\circ }$)
We have given that $ABCD$ is a cyclic quadrilateral, therefore, the sum of opposite sides of the quadrilateral is ${180^\circ }$.
$\therefore \angle BCD + \angle BAD = {180^\circ }$
$ \Rightarrow \angle BCD + {90^\circ } = {180^\circ }$
$ \Rightarrow \angle BCD = {180^\circ } - {90^\circ }$
$ \Rightarrow \angle BCD = {90^\circ }$ ….. (2)
Similarly, we can also show that $\angle ADC = {90^\circ }$ as follows:
Since $AC$ is the diameter and hence arc$ABC$ is in the semi circle.
$\therefore \angle ABC = {90^\circ }$ ….. (3) (Angle in a semicircle is ${90^\circ }$)
We have given that $ABCD$ is a cyclic quadrilateral, therefore, the sum of opposite sides of the quadrilateral is ${180^\circ }$.
$\therefore \angle ABC + \angle ADC = {180^\circ }$
$ \Rightarrow {90^\circ } + \angle ADC = {180^\circ }$
$ \Rightarrow \angle ADC = {180^\circ } - {90^\circ }$
$ \Rightarrow \angle ADC = {90^\circ }$ ….. (4)
From (1), (2), (3) and (4), we get-
$\angle A = \angle B = \angle C = \angle D = {90^\circ }$
Since $\angle A = \angle C$ and $\angle B = \angle D$, i.e., both pairs of opposite angles are equal .
Therefore, $ABCD$ is a parallelogram.
Also, all angles are ${90^\circ }$.
So, $ABCD$ is a parallelogram with one angle ${90^\circ }$.

Therefore, $ABCD$ is a rectangle.


Note:
It may be noted that a cyclic quadrilateral is one whose sum of opposite sides of is ${180^\circ }$. So a quadrilateral $ABCD$ is a cyclic quadrilateral if: $\angle A + \angle C = {180^\circ }$ and $\angle B + \angle D = {180^\circ }$.