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"The sum of the squares of the diagonals is equal to the sum of the squares of the four sides of a parallelogram."

I find this property very useful while solving different problems on Quadrilaterals & Polygon,so I am very inquisitive about a intuitive proof of this property.

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    Did you manage to do it using Pythagoras now? If not, look at [this picture](http://i.stack.imgur.com/qpjIw.png) and calculate: \begin{align*} b^2 &= x^2 + h^2 \\ e^2 &= (a-x)^2 + h^2 = a^2 -2ax+x^2 + h^2 \\ f^2 &= (a+x)^2 + h^2 = a^2 +2ax+x^2 + h^2 \end{align*} So $2a^2 + 2b^2 = 2a^2 + 2x^2 + 2h^2$ and $e^2 + f^2 = 2a^2 + 2x^2 + 2h^2$, hence $2a^2 + 2b^2 = e^2 + f^2$.2011-09-04

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Let two sides of the parallelogram correspond to the vectors $\vec a$ and $\vec b$. Then the diagonals correspond to the vectors $\vec a+\vec b$ and $\vec a-\vec b$, and

$(\vec a+\vec b)^2+(\vec a-\vec b)^2=2\vec a^2+2\vec b^2\;,$

which is the desired identity between the sums of the squares of the diagonals and the sides.

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    ...and just in case it isn't obvious, in this answer $\vec{x}^2$ is used as an abbreviation for $\vec{x} \cdot \vec{x}$ (inner product).2011-09-04