Question

Prove that the rhombus with equal diagonals is a square.

Answer 1

Hint:Draw a figure of rhombus when diagonals are equal. In rhombus, the diagonals bisect each other at right angles. Prove that the vertices are also right angles.

Complete step by step answer:
From the figure, let us say that ABCD is a rhombus.

Let AC and BD be the diagonals of the rhombus. It is said that the rhombus has equal diagonals, i.e. AC = BD.
We know that diagonals of rhombus bisect each other. Thus we can say that AC = BD.
If AC = BD, then
AO = BO = CO = DO…..(1)
Consider the point where the diagonals intersect at O.
Let us consider the $$\mathrm{\Delta }AOB$$.
From equation (1) we can say that AO = BO and $$\mathrm{\angle }AOB={90}^{\circ }$$.
In rhombus, the diagonals bisect each other at right angles. Each diagonal cuts the other into two equal parts and the angle where they cross is always a right angle.
$$\therefore \mathrm{\angle }AOB={90}^{\circ }$$.
Thus we can say that $$\mathrm{\angle }OAB=\mathrm{\angle }OBA={\displaystyle \frac{{90}^{\circ }}{2}}={45}^{\circ }$$
Similarly in $$\mathrm{\Delta }AOD$$, $$\mathrm{\angle }OAD=\mathrm{\angle }ODA={\displaystyle \frac{{90}^{\circ }}{2}}={45}^{\circ }$$
Thus from the figure,
$$\mathrm{\angle }A=\mathrm{\angle }OAB+\mathrm{\angle }OAD={45}^{\circ }+{45}^{\circ }={90}^{\circ }$$
Similarly, from the figure,
$$\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$$
$$\therefore$$All the angles, $$\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$$
All the sides are also equal, i.e. AB = BC = CD = DA.
$$\therefore$$Quadrilateral ABCD is a square.
Therefore, we proved that a rhombus with equal diagonals is a square.

Note:In case of a square, we know the general properties, that all sides of a square are equal in length. Thus the diagonals of a square are equal in length and diagonals bisect each other. The angles are $${90}^{\circ }$$. Thus the square has 4 congruent sides and 4 right angles.