# Using vector method, prove that if the diagonals of a parallelogram are equal in length, then it is a rectangle.

Toolbox:
• To prove that the parallelogram is a rectangle prove that the adjacent sides are $\perp$
• If $\overrightarrow a.\overrightarrow b=0,\:then\:\overrightarrow a\perp\overrightarrow b$
• $|\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b$
• $|\overrightarrow a-\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2-2\overrightarrow a.\overrightarrow b$
Let OABC be a parallelogram woth O as origin OB and AC as diagonals.
Let $\overrightarrow {OA}=\overrightarrow a,\:\overrightarrow {OB}=\overrightarrow b$ and $\overrightarrow {OC}=\overrightarrow c$
The diagonals are $\overrightarrow {OB}=\overrightarrow b\:and\:\overrightarrow {AC}=\overrightarrow c-\overrightarrow a$
Adjacent sides are $\overrightarrow {OA}=\overrightarrow a\:and\:\overrightarrow {OC}=\overrightarrow c$
Also we know from parallelogram law of addition that the two diagonals of the parallelogram are given by
$\overrightarrow {OB}=\overrightarrow a+\overrightarrow c$ and $\overrightarrow {AC}=\overrightarrow c-\overrightarrow a$
By the given condition the diagonals are of equal length.
$\Rightarrow\:|\overrightarrow {OB}|=|\overrightarrow {AC}|$
$\Rightarrow\:|\overrightarrow c+\overrightarrow a|=|\overrightarrow c-\overrightarrow b|$
$\Rightarrow\:|\overrightarrow c+\overrightarrow a|^2=|\overrightarrow c-\overrightarrow b|^2$
$\Rightarrow\:|\overrightarrow c|^2+|\overrightarrow a|^2+2\overrightarrow a.\overrightarrow c=|\overrightarrow c|^2+|\overrightarrow a|^2-2\overrightarrow a.\overrightarrow c$
$\Rightarrow \:2\overrightarrow a.\overrightarrow c=-2\overrightarrow a.\overrightarrow c$
$\Rightarrow \overrightarrow a.\overrightarrow c=0$
$\Rightarrow a\:is\:\perp\:to\:\overrightarrow c$ where $\overrightarrow a$ and $\overrightarrow c$ are adjacent sides.
We know that in a parallelogram if the adjacent sides are $\perp$ then it is a rectangle.
Hence proved.
edited Apr 24, 2013