Question

A quadrilateral whose diagonals are equal and bisect each other at right angles is called a:
$$\begin{gathered} \left(a\right)\text{Rhombus} \\ \left(b\right)\text{Square} \\ \left(c\right)\text{Rectangle} \\ \left(d\right)\text{Trapezium} \end{gathered}$$

Answer 1

Hint: In this question, we have to use the given property of quadrilateral and then prove any unique property of quadrilateral. So, we can easily identify the correct option.

Complete step-by-step answer:

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at o.

Given, the diagonals of ABCD are equal and bisect each other at right angles.

Therefore, AC = BD, OA = OC, OB = OD, and $\mathrm{\angle }AOB=\mathrm{\angle }BOC=\mathrm{\angle }COD=\mathrm{\angle }AOD={90}^0$

Now, In $△AOB$ and $△COD$
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
$\mathrm{\angle }AOB=\mathrm{\angle }COD={90}^0$ (Vertically opposite angles)
So, $△AOB\cong △COD$ (SAS congruence rule)
Hence AB = CD (By CPCT) ...........(1)
And, $\mathrm{\angle }OAB=\mathrm{\angle }OCD$ (By 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.

So, $AB\parallel CD............\left(2\right)$
From (1) and (2) equation,
ABCD is a parallelogram.

In $△AOD$ and $△COD$
AO = CO (Diagonals bisect each other)
OD = OD (Common)
$\mathrm{\angle }AOD=\mathrm{\angle }COD={90}^0$
So, $△AOD\cong △COD$ (SAS congruence rule)
Hence AD = DC ……….. (3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)
So, AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.
In $△ADC$ and $△BCD$
AD = BC (Already proved)
AC = BD (Given)
CD = CD (Common)

So, $△ADC\cong △BCD$ (SSS Congruence rule)
Hence, $\mathrm{\angle }ADC=\mathrm{\angle }BCD$ (By CPCT)

However, $\mathrm{\angle }ADC+\mathrm{\angle }BCD={180}^0$ (Co-interior angles)
$$\begin{gathered} \Rightarrow \mathrm{\angle }ADC+\mathrm{\angle }ADC={180}^0 \\ \Rightarrow 2\mathrm{\angle }ADC={180}^0 \\ \Rightarrow \mathrm{\angle }ADC={90}^0 \end{gathered}$$
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 ${90}^0$ . Therefore, ABCD is a square.

So, the correct option is (b).

Note: Whenever we face such types of problems we use some important points. Like first we prove that quadrilateral ABCD is a parallelogram and all sides of quadrilateral are equal then prove one of the interior angles of quadrilateral ABCD is a right angle. So, according to this property it is proved that the quadrilateral is Square.