91Ó°ÊÓ

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

Consider the following rhombus,

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

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

And

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

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

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

Since you know that,

$\left|\stackrel{→}{\mathrm{A}}\right|=\left|\stackrel{→}{\mathrm{B}}\right|$

Therefore,

$\stackrel{→}{{\mathrm{d}}_1}×{\stackrel{→}{\mathrm{d}}}_2=0$

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

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.