# Prove by the vector method If the diagonals of a parallelogram are equal than it is a rectangle.

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• If $\overrightarrow a \perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$ and for nonzero vectors if $\overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a \perp \overrightarrow b.$
• For any two vectors $\hat a \: and \: \hat b$ $(\hat a + \hat b)^2=(\hat a)^2+2\hat a.\hat b+(\hat b)^2=a^2+2\hat a.\hat b+b^2$ $(\hat a-\hat b)^2=a^2-2\hat a.\hat b+b^2$ $(\hat a+\hat b).(\hat a-\hat b)=a^2-b^2$
• By $\Delta$ law of vectors if $\overrightarrow a+\overrightarrow b=\overrightarrow c \: or \: \overrightarrow a+\overrightarrow b=-\overrightarrow c$ then the vectors form the sides of a $\Delta$
Let ABCD be the parallelogram with sides $\overrightarrow {AB}=\overrightarrow a\: and \overrightarrow {AD}=\overrightarrow b$ The diagonals are $\overrightarrow {AC}\: and \:\: \overrightarrow {BD}$
$\overrightarrow {AC}=\overrightarrow {AB}+\overrightarrow {BC}=\overrightarrow {AB}+\overrightarrow {AD}=\overrightarrow a+\overrightarrow b$
$\overrightarrow {BD}=\overrightarrow {BA}+\overrightarrow {AD}=\overrightarrow {AD}-\overrightarrow {AB}=\overrightarrow b-\overrightarrow a$
Since the diagonals are equal, $|\overrightarrow {AC}|=|\overrightarrow {BD}| \: or \: |\overrightarrow a+\overrightarrow b|=|\overrightarrow b-\overrightarrow a|$
$\therefore (\overrightarrow a+\overrightarrow b)^2=(\overrightarrow b-\overrightarrow a)^2$
$\overrightarrow a^2+2\overrightarrow a.\overrightarrow b+\overrightarrow b^2=\overrightarrow b^2-2\overrightarrow b.\overrightarrow a+\overrightarrow a^2$
$\Rightarrow 2\overrightarrow a.\overrightarrow b=-2\overrightarrow a.\overrightarrow b\: or \: \overrightarrow a.\overrightarrow b=0$
Since $\overrightarrow a\: and \: \overrightarrow b$ are nonzero, it follows that $\overrightarrow a \perp \overrightarrow b. \: \therefore$ ABCD is a rectangle.

edited Jun 20, 2013