Question

Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer 1

Let OACB be a quadrilateral such that diagonals OC and AB bisect each other at 90º.

Taking O as the origin, let the poisition vectors of A and B be $\overrightarrow{a}$ and $\overrightarrow{b}$, respectively. Then,

$\overrightarrow{\mathrm{OA}}=\overrightarrow{a}$ and $\overrightarrow{\mathrm{OB}}=\overrightarrow{b}$
Position vector of mid-point of AB, $\overrightarrow{\mathrm{OE}}=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$

∴ Position vector of C, $\overrightarrow{\mathrm{OC}}=\overrightarrow{a}+\overrightarrow{b}$

By the triangle law of vector addition, we have

$$\begin{gathered} \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{b}-\overrightarrow{a} \end{gathered}$$

Since $\overrightarrow{\mathrm{AB}}\perp \overrightarrow{\mathrm{OC}}$,

$$\begin{gathered} \Rightarrow \overrightarrow{\mathrm{AB}}.\overrightarrow{\mathrm{OC}}=0 \\ \Rightarrow \left(\overrightarrow{b}-\overrightarrow{a}\right).\left(\overrightarrow{a}+\overrightarrow{b}\right)=0 \\ \Rightarrow {\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{a}\right|}^2=0 \\ \Rightarrow {\left|\overrightarrow{a}\right|}^2={\left|\overrightarrow{b}\right|}^2 \\ \Rightarrow \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right| \\ \Rightarrow \mathrm{OA}=\mathrm{OB} \end{gathered}$$

In a quadrilateral if diagonals bisects each other at right angle and adjacent sides are equal, then it is a rhombus.