In a kite (four-sided figure made up of two pairs of equal adjacent sides), the diagonals are perpendicular.

For two vectors $\overrightarrow{\mathbf{x}}$and$\mathbf{}\overrightarrow{\mathbf{y}}$ ,

role="math" localid="1658991640641" $$\begin{gathered} {\mathbf{|}\overrightarrow{\mathbf{x}}\mathbf{-}\mathbf{}\overrightarrow{\mathbf{y}}\mathbf{|}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\overrightarrow{\mathbf{x}}\mathbf{\times }\overrightarrow{\mathbf{x}}\mathbf{)}\mathbf{-}\mathbf{\left(}\mathbf{2}\overrightarrow{\mathbf{x}}\overrightarrow{\mathbf{y}}\mathbf{\right)}\mathbf{+}\mathbf{(}\overrightarrow{\mathbf{y}}\mathbf{\times }\overrightarrow{\mathbf{y}}\mathbf{)} \\ \mathbf{=}{\mathbf{\left|}\overrightarrow{\mathbf{x}}\mathbf{\right|}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\overrightarrow{\mathbf{x}}\overrightarrow{\mathbf{y}}\mathbf{+}{\mathbf{\left|}\overrightarrow{\mathbf{y}}\mathbf{\right|}}^{\mathbf{2}} \end{gathered}$$

Let Q A B C be a kite in which adjacent sides O A and O C are equal. Other two adjacent sides C B and A B also are equal.

Let $\overrightarrow{a},\overrightarrow{b}$and $\overrightarrow{c}$are the position vectors of points A, B, and C respectively as shown in figure here.

Therefore,

$$\begin{gathered} \mathrm{OA}=\mathrm{OC} \\ \Rightarrow \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{OC}} \\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{c}}\right|\dots ..\left(1\right) \\ \mathrm{AB}=\mathrm{CB} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{CB}} \\ \Rightarrow |\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}|\dots ..\left(2\right) \end{gathered}$$

The vector of diagonal OB is $\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}},$and vector of diagonal CA is $\overrightarrow{\mathrm{CA}}=(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}})$.

The scalar product of vectors $\overrightarrow{OB}$ and $\overrightarrow{CA}$

$$\begin{gathered} \overrightarrow{OB}\times \overrightarrow{CA}=\overrightarrow{b}\times (\overrightarrow{a}-\overrightarrow{c}) \\ \overrightarrow{OB}\times \overrightarrow{CA}=\overrightarrow{b}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{c}\dots ..\left(3\right) \end{gathered}$$

Squaring both the sides in equation (2)

$$\begin{gathered} |\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}{|}^2=|\overrightarrow{\mathrm{b}}-\mathrm{c}⃗{|}^2 \\ \Rightarrow |\overrightarrow{\mathrm{b}}{|}^2-2\mathrm{b}⃗\times \mathrm{a}⃗+|\mathrm{a}⃗{|}^2=|\mathrm{b}⃗{|}^2-2\overrightarrow{\mathrm{b}}\times \mathrm{c}⃗+|\mathrm{c}⃗{|}^2 \end{gathered}$$

But as known that,

$\left|\overrightarrow{c}\right|=\left|\overrightarrow{a}\right|$

So,

role="math" localid="1658993663930" $\overrightarrow{b}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{c}$ ….. (4)

Substituting the above equation in result (3), and you have

$$\begin{gathered} \overrightarrow{OB}\times \overrightarrow{CA}=\overrightarrow{b}\times \overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{c} \\ =0 \end{gathered}$$

Since, the scalar product of two vectors $\overrightarrow{OB}$ and $\overrightarrow{CA}$ is zero, hence, they are perpendicular to each other.

Thus, the diagonals are perpendicular in the kite.