Question

Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.

Answer 1

ABCD is a rectangle. Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA, respectively.

Now,

$\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{BQ}}=\frac{1}{2}\overrightarrow{\mathrm{AB}}+\frac{1}{2}\overrightarrow{\mathrm{BC}}=\frac{1}{2}\left(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}\right)=\frac{1}{2}\overrightarrow{\mathrm{AC}}$ .....(1)

$\overrightarrow{\mathrm{SR}}=\overrightarrow{\mathrm{SD}}+\overrightarrow{\mathrm{DR}}=\frac{1}{2}\overrightarrow{\mathrm{AD}}+\frac{1}{2}\overrightarrow{\mathrm{DC}}=\frac{1}{2}\left(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DC}}\right)=\frac{1}{2}\overrightarrow{\mathrm{AC}}$ .....(2)

From (1) and (2), we have

$\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}}$

So, the sides PQ and SR are equal and parallel. Thus, PQRS is a parallelogram.

Now,

$$\begin{gathered} {\left|\overrightarrow{\mathrm{PQ}}\right|}^2=\overrightarrow{\mathrm{PQ}}.\overrightarrow{\mathrm{PQ}} \\ \Rightarrow {\left|\overrightarrow{\mathrm{PQ}}\right|}^2=\left(\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{BQ}}\right).\left(\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{BQ}}\right) \\ \Rightarrow {\left|\overrightarrow{\mathrm{PQ}}\right|}^2={\left|\overrightarrow{\mathrm{PB}}\right|}^2+2\overrightarrow{\mathrm{PB}}.\overrightarrow{\mathrm{BQ}}+{\left|\overrightarrow{\mathrm{BQ}}\right|}^2 \\ \Rightarrow {\left|\overrightarrow{\mathrm{PQ}}\right|}^2={\left|\overrightarrow{\mathrm{PB}}\right|}^2+0+{\left|\overrightarrow{\mathrm{BQ}}\right|}^2\left(\overrightarrow{\mathrm{PB}}\perp \overrightarrow{\mathrm{BQ}}\right) \\ \Rightarrow {\left|\overrightarrow{\mathrm{PQ}}\right|}^2={\left|\overrightarrow{\mathrm{PB}}\right|}^2+{\left|\overrightarrow{\mathrm{BQ}}\right|}^2.....\left(3\right) \end{gathered}$$

Also,

$$\begin{gathered} {\left|\overrightarrow{\mathrm{PS}}\right|}^2=\overrightarrow{\mathrm{PS}}.\overrightarrow{\mathrm{PS}} \\ \Rightarrow {\left|\overrightarrow{\mathrm{PS}}\right|}^2=\left(\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{AS}}\right).\left(\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{AS}}\right) \\ \Rightarrow {\left|\overrightarrow{\mathrm{PS}}\right|}^2={\left|\overrightarrow{\mathrm{PA}}\right|}^2+2\overrightarrow{\mathrm{PA}}.\overrightarrow{\mathrm{AS}}+{\left|\overrightarrow{\mathrm{AS}}\right|}^2 \\ \Rightarrow {\left|\overrightarrow{\mathrm{PS}}\right|}^2={\left|\overrightarrow{\mathrm{PB}}\right|}^2+0+{\left|\overrightarrow{\mathrm{BQ}}\right|}^2\left(\overrightarrow{\mathrm{PA}}\perp \overrightarrow{\mathrm{AS}}\right) \\ \Rightarrow {\left|\overrightarrow{\mathrm{PS}}\right|}^2={\left|\overrightarrow{\mathrm{PB}}\right|}^2+{\left|\overrightarrow{\mathrm{BQ}}\right|}^2.....\left(4\right) \end{gathered}$$

From (3) and (4), we have

$$\begin{gathered} {\left|\overrightarrow{\mathrm{PQ}}\right|}^2={\left|\overrightarrow{\mathrm{PS}}\right|}^2 \\ \Rightarrow \left|\overrightarrow{\mathrm{PQ}}\right|=\left|\overrightarrow{\mathrm{PS}}\right| \end{gathered}$$

So, the adjacent sides of the parallelogram are equal. Hence, PQRS is a rhombus.