Prove that the rhombus with equal diagonals is a square.

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 \[\Delta AOB\].
From equation (1) we can say that AO = BO and \[\angle AOB={{90}^{\circ }}\].
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.
\[\therefore \angle AOB={{90}^{\circ }}\].
Thus we can say that \[\angle OAB=\angle OBA=\dfrac{{{90}^{\circ }}}{2}={{45}^{\circ }}\]
Similarly in \[\Delta AOD\], \[\angle OAD=\angle ODA=\dfrac{{{90}^{\circ }}}{2}={{45}^{\circ }}\]
Thus from the figure,
\[\angle A=\angle OAB+\angle OAD={{45}^{\circ }}+{{45}^{\circ }}={{90}^{\circ }}\]
Similarly, from the figure,
\[\angle B=\angle C=\angle D={{90}^{\circ }}\]
\[\therefore \]All the angles, \[\angle A=\angle B=\angle C=\angle D={{90}^{\circ }}\]
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.