Question

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer 1

Let ABCD be a parallelogram such that AC and BD are its two diagonals. Taking A as the origin, let the position vectors of B and D be $\overrightarrow{b}$ and $\overrightarrow{d}$, respectively. Then,

$\overrightarrow{\mathrm{AB}}=\overrightarrow{b}$ and $\overrightarrow{\mathrm{AD}}=\overrightarrow{d}$

Using triangle law of vector addition, we have

$$\begin{gathered} \overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DB}}=\overrightarrow{\mathrm{AB}} \\ \Rightarrow \overrightarrow{\mathrm{DB}}=\overrightarrow{b}-\overrightarrow{d} \end{gathered}$$

In ∆ABC,

$\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}=\overrightarrow{b}+\overrightarrow{d}$

Now,

$$\begin{gathered} {\left|\overrightarrow{\mathrm{AB}}\right|}^2+{\left|\overrightarrow{\mathrm{BC}}\right|}^2+{\left|\overrightarrow{\mathrm{CD}}\right|}^2+{\left|\overrightarrow{\mathrm{DA}}\right|}^2 \\ ={\left|\overrightarrow{\mathrm{AB}}\right|}^2+{\left|\overrightarrow{\mathrm{AD}}\right|}^2+{\left|-\overrightarrow{\mathrm{AB}}\right|}^2+{\left|-\overrightarrow{\mathrm{AD}}\right|}^2 \\ =2{\left|\overrightarrow{\mathrm{AB}}\right|}^2+2{\left|\overrightarrow{\mathrm{AD}}\right|}^2 \\ =2{\left|\overrightarrow{b}\right|}^2+2{\left|\overrightarrow{d}\right|}^2.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} {\left|\overrightarrow{\mathrm{DB}}\right|}^2+{\left|\overrightarrow{\mathrm{AC}}\right|}^2 \\ ={\left|\overrightarrow{b}-\overrightarrow{d}\right|}^2+{\left|\overrightarrow{b}+\overrightarrow{\mathit{d}}\right|}^2 \\ =\left(\overrightarrow{b}-\overrightarrow{\mathit{d}}\right).\left(\overrightarrow{\mathit{b}}-\overrightarrow{\mathit{d}}\right)+\left(\overrightarrow{\mathit{b}}+\overrightarrow{d}\right).\left(\overrightarrow{\mathit{b}}+\overrightarrow{d}\right) \\ ={\left|\overrightarrow{b}\right|}^2-2\overrightarrow{\mathit{b}}\mathit{.}\overrightarrow{\mathit{d}}+{\left|\overrightarrow{\mathit{d}}\right|}^2+{\left|\overrightarrow{\mathit{b}}\right|}^2+2\overrightarrow{\mathit{b}}\mathit{.}\overrightarrow{\mathit{d}}+{\left|\overrightarrow{d}\right|}^2 \\ =2{\left|\overrightarrow{b}\right|}^2+2{\left|\overrightarrow{d}\right|}^2.....\left(2\right) \end{gathered}$$

From (1) and (2), we have

${\left|\overrightarrow{\mathrm{AB}}\right|}^2+{\left|\overrightarrow{\mathrm{BC}}\right|}^2+{\left|\overrightarrow{\mathrm{CD}}\right|}^2+{\left|\overrightarrow{\mathrm{DA}}\right|}^2={\left|\overrightarrow{\mathrm{DB}}\right|}^2+{\left|\overrightarrow{\mathrm{AC}}\right|}^2$