# Using vector method, show that the diagonals of a Rhombus bisect each other at right angles.

Toolbox:
• Rhombus is a parallelogram with all the four sides equal.
• If $\overrightarrow a.\overrightarrow b=0$ then $\overrightarrow a\:is\:\perp\:to\:\overrightarrow b$
• $(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2-|\overrightarrow b|^2$
• Parallelogram law of addition: In a parallelogram ABCD $\overrightarrow {AC}=\overrightarrow {AB}+\overrightarrow {AD}\:and\:\overrightarrow {DB}=\overrightarrow {AB}-\overrightarrow {AD}$
Let ABCD be a rhombus.
$\Rightarrow |\overrightarrow {AB}|=|\overrightarrow {BC}|=|\overrightarrow {DC}|=|\overrightarrow {AD}|$
The diagonals are $\overrightarrow {AC}\:and\:\overrightarrow {DB}$
Since Rhombus is a parallelogram we know that in a parallelogram
the diagonals bisect each other.
We have to prove that the diagonals bisect perpendicularly.
i.e., we have to prove that $\overrightarrow {AC}\:is\:\perp\:to\:\overrightarrow {DB}$
From parallelogram law of addition we know that
$\overrightarrow {AC}=\overrightarrow {AB}+\overrightarrow {AD}\:and\:\overrightarrow {DB}=\overrightarrow {AB}-\overrightarrow {AD}$
To prove that $\overrightarrow {AC}\:is\:\perp\:to\:\overrightarrow {DB}$
$\Rightarrow prove \:\overrightarrow {AC}.\overrightarrow {DB}=0$
$\overrightarrow {AC}.\overrightarrow {DB}=(\overrightarrow {AB}+\overrightarrow {AD}).(\overrightarrow {AB}-\overrightarrow {AD})$
We know that $(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2-|\overrightarrow b|^2$
$\Rightarrow\:\overrightarrow {AC}.\overrightarrow {DB}=|\overrightarrow {AB}|^2-|\overrightarrow {AD}|^2$
Since ABCD is a Rhombus $|\overrightarrow {AB}|=|\overrightarrow {AD}|$
$\Rightarrow \overrightarrow {AC}.\overrightarrow {DB}=0$
$\Rightarrow \overrightarrow {AC}\:is\:\perp\:to\:\overrightarrow {DB}$
$\Rightarrow$ the diagonals of a Rhombus are perpendicular bisectors.
Hence proved.