Question

The diagonals of a rhombus (four-sided figure with all sides of equal length) are perpendicular and bisect each other.

Short answer

The diagonals of a rhombus are orthogonal and bisect each other.

Step-by-step solution

Concept and formula used:

A rhombus is a special case of a parallelogram, and it is a four-sided quadrilateral.

In a rhombus, opposite sides are parallel and the opposite angles are equal. Moreover,all the sides of a rhombus are equal in length, and the diagonals bisect each other at right angles.

Corresponding parts of congruent triangles are congruent, so all 4 angles (the ones in the middle) are congruent

To prove the diagonals of a rhombus are orthogonal and bisect each other.

Consider the following rhombus,

From the figure, the vectors $\overrightarrow{A}$and $\overrightarrow{\mathrm{B}}$are equal. Therefore,

$\left|\overrightarrow{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{B}}\right|$,

And

$$\begin{gathered} \overrightarrow{{\mathrm{d}}_1}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}} \\ \overrightarrow{{\mathrm{d}}_2}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}} \end{gathered}$$

Take a cross product of $\overrightarrow{d_1}$and $\overrightarrow{d_2}$, and you get

$$\begin{gathered} \overrightarrow{{\mathrm{d}}_1}\times \overrightarrow{{\mathrm{d}}_2}=\left(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\right)\times \left(\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\right) \\ =\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}} \\ =\left|\overrightarrow{\mathrm{B}}\right|-{\left|\overrightarrow{\mathrm{A}}\right|}^2 \end{gathered}$$

Since you know that,

$\left|\overrightarrow{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{B}}\right|$

Therefore,

$\overrightarrow{{\mathrm{d}}_1}\times {\overrightarrow{\mathrm{d}}}_2=0$

And so ${\overrightarrow{d}}_1$ is perpendicular to ${\overrightarrow{d}}_2$.

Note that the rhombus is a special case of parallelogram, where all the sides have the same length. And proved that for any parallelogram, its diagonals bisect each other.