Question

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

Answer 1

Let OABC be a rhombus, whose diagonals OB and AC intersect at D. Suppose O is the origin.

Let the position vector of A and C be $\overrightarrow{a}$ and $\overrightarrow{c}$, respectively. Then,

$\overrightarrow{\mathrm{OA}}=\overrightarrow{a}$ and $\overrightarrow{\mathrm{OC}}=\overrightarrow{c}$

In ∆OAB,

$\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=\overrightarrow{a}+\overrightarrow{c}$ $\left(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OC}}\right)$

Position vector of mid-point of $\overrightarrow{\mathrm{OB}}=\frac{1}{2}\left(\overrightarrow{a}+\overrightarrow{c}\right)$

Position vector of mid-point of $\overrightarrow{\mathrm{AC}}=\frac{1}{2}\left(\overrightarrow{a}+\overrightarrow{c}\right)$ (Mid-point formula)

So, the mid-points of OB and AC coincide. Thus, the diagonals OB and AC bisect each other.

Now,

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

Hence, the diagonals OB and AC are perpendicular to each other.

Thus, the diagonals of a rhombus are perpendicular bisectors of each other.