Prove by vector method that if the diagonals of a parallelogram are equal, then the parallelogram is a rectangle.
Let $ABCD$ be a parallelogram.
Let $\overrightarrow{AB} = \overrightarrow{a}$, $\;$ $\overrightarrow{BC} = \overrightarrow{b}$
Then $\overrightarrow{CD} = - \overrightarrow{a}$, $\;$ $\overrightarrow{DA} = - \overrightarrow{b}$
Also, $\left|\overrightarrow{AB}\right| = \left|\overrightarrow{CD}\right|$; $\;$ $\left|\overrightarrow{BC}\right| = \left|\overrightarrow{DA}\right|$ $\;\;\; \cdots \; (1)$
Diagonal $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{a} + \overrightarrow{b}$
Diagonal $\overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB} = \overrightarrow{a} - \overrightarrow{b}$
Given: $\;$ $\left|\overrightarrow{AC}\right| = \left|\overrightarrow{DB}\right|$
i.e. $\;$ $\left|\overrightarrow{a} + \overrightarrow{b}\right| = \left|\overrightarrow{a} - \overrightarrow{b}\right|$
i.e. $\;$ $\left|\overrightarrow{a} + \overrightarrow{b}\right|^2 = \left|\overrightarrow{a} - \overrightarrow{b}\right|^2$
i.e. $\;$ $\left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{b}\right|^2 + 2 \left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right| = \left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{b}\right|^2 - 2 \left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right|$
i.e. $\;$ $4 \left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right| = 0$
i.e. $\;$ $\left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right| = 0$
$\implies$ $\overrightarrow{a} \perp \overrightarrow{b}$ (i.e. the adjacent sides are perpendicular to each other) $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ We have from $(1)$ and $(2)$, if the diagonals of a parallelogram are equal, then the parallelogram is a rectangle.