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# Prove by vector method , The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of the sides.

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• 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
Now $\overrightarrow {AC}= \overrightarrow {AB}+ \overrightarrow {BC}$
$\overrightarrow {BD}= \overrightarrow {BA}+ \overrightarrow {AD}$
$AC^2=AC^2 = ( \overrightarrow {AB}+ \overrightarrow {BC})^2= \overrightarrow {AB}^2+2 \overrightarrow {AB}. \overrightarrow {BC}+ \overrightarrow {BC}^2$
$= AB^2+2 \overrightarrow {AB}. \overrightarrow {AD}+ BC^2$ (i)
$\overrightarrow {BD}^2 = BD^2=( \overrightarrow {BA}+ \overrightarrow {AD})^2=( \overrightarrow {AD}- \overrightarrow {AB})^2= \overrightarrow {AD}^2-1 \overrightarrow {AD}. \overrightarrow {AB}+ \overrightarrow {AB}^2$
$= AD^2-2\overrightarrow {AB}.\overrightarrow {AD}+\overrightarrow {AB}^2$ (ii)
$AC^2=BD^2=AB^2+\not2\overrightarrow {AB}.\overrightarrow {AD}+BC^2+AD^2-\not2\overrightarrow {AD}.\overrightarrow {AB}+AB^2$
$= AB^2+BC^2+AD^2+AB^2$ Hence proved
edited Jun 6, 2013 by meena.p