# Ex.8.1 Q5 Quadrilaterals Solution - NCERT Maths Class 9

## Question

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

## Text Solution

**What is known?**

The diagonals of a quadrilateral are equal and bisect each other at right angle.

**What is unknown?**

How we can show that it is a square.

**Reasoning:**

We have to show that given quadrilateral is a parallelogram in which all sides are equal and one of its interior angles is .

**Steps:**

Let us consider a quadrilateral \(ABCD\) in which the diagonals \(AC\) and \(BD\) intersect each other at \(O\). It is given that the diagonals of \(ABCD\) are equal and bisect each other at right angles.

Therefore,

\({\rm{AC }} = {\rm{ BD}},{\rm{ OA }} = {\rm{ OC}},{\rm{ OB }} = {\rm{ OD}},\) and

\(\angle \,{\rm{AOB }} \!=\!\angle {\rm{BOC }} \!=\! \angle {\rm{COD }} \!=\! \angle {\rm{AOD }} \!=\! {\rm{ 9}}{0^0}\)

To prove \(ABCD\) is a square

We have to prove that \(ABCD\) is a parallelogram,in which

\(AB = BC = CD = AD\), and one of its interior angles is \(90^{0}\) .

In \( { \Delta{AOD} }{ {\: \rm and } \:\Delta {COD}}\)

\[\begin{align}{{AO}}&={CO}\\ \text{(Diagonals }& \text { bisect each other)} \\\\ \angle {AOD}&=\angle {COD}\\( \text{ Given that} &\text{ each is } 90^{\circ}) \\\\ {{OD}}&={{OD}\quad( \text{ Common })}\\ \therefore \Delta {AOD} &\cong \Delta {COD}\\ \text{(SAS}& \text { congruence rule)} \\\\ \therefore {AD}&={{DC}}\quad\dots(3)\end{align}\]

\[\rm {And,}\angle \mathrm{OAB}=\angle \mathrm{OCD}(\mathrm{By} \,\text {CPCT})\]

However, these are alternate interior angles for line \(AB\) and \(CD\) and alternate interior angles are equal to each other only when the two lines are parallel.

\(AB \;|| \;CD \)... (2)

From Equations (1) and (2), we obtain \(ABCD\) is a parallelogram.

In \(\Delta AOD\) and \(\Delta {COD}\)

\[\begin{align} {{AO}}&={CO}\\ \text{(Diagonals }& \text { bisect each other)} \\ \\\ {\angle {AOD}}&=\angle {COD}\\( \text{ Given that} &\text{ each is } 90^{\circ})\\\\ {{OD}}&={{OD}\,\,( \text{ Common })}\\ \therefore \Delta {AOD} &\cong \Delta {COD}\\ \text{(SAS}& \text { congruence rule)} \\\\ \therefore {AD}&={{DC}}\qquad\dots(3)\end{align}\]

However, \(AD = BC\) and \(AB = CD\)

(Opposite sides of parallelogram \(ABCD\))

\(AB = BC = CD = DA\)

Therefore, all the sides of quadrilateral \(ABCD\) are equal to each other.

In \( \Delta {ADC}\) and \( \Delta {BCD} ,\)

\[\begin{align}\\ {{AD}}&={{BC}\; ( \text{ Already proved })} \\ {{AC}}&={{BD} \;\text{ (Given) }} \\ {{DC}}&={{CD}\; ( \text{ Common })}\\{\therefore \Delta {ADC} }&\cong \Delta {BCD}\\ \text{(SAS}& \text { congruence rule)} \\\\ {\therefore\angle {ADC}}&={\angle {BCD}\;(\text{By} \,\text{CPCT})}\end{align}\]

However,

\[\begin{align} \angle {ADC}\!+\angle {BCD}&=180^{\circ}\\ \text{ (Co-interior }& \text {angles) } \\\\ {\qquad \angle {ADC}+\angle {ADC}}&={180^{\circ}}\\{2 \angle {ADC}}&={180^{\circ}} \\ {\therefore \angle {ADC}}&={90^{\circ}}\end{align}\]

One of the interior angles of quadrilateral \(ABCD\) is a right angle.

Thus, we have obtained that \(ABCD\) is a parallelogram, \(AB = BC = CD = AD\) and one of its interior angles is \(90^{\circ}\) .

Therefore, \(ABCD\) is a square.