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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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    You could try splitting the parallelogram into two triangles and apply the law of cosines to each of the two triangles...2011-09-04
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    And I suppose cosine law would count as non-intuitive :-). Added: @J.M. I had typed this before seeing your comment, not as a comment on your comment!2011-09-04
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    @J. M.:That could work,but as this problem is most often taught in elementary classes (prior to law of sines and cosines),so there could be more elementary/elegant + intuitive proof :)2011-09-04
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    Is the case of a rectangle (read Pythagoras) intuitive enough?2011-09-04
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    @Theo Buehler:I did thought about using Pythagoras but I am not sure if we could use it on a Rhomboid.2011-09-04
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    To finish @Theo's suggestion: remember that you can snip off a right triangle with one side being the height of the parallelogram and rejoin that to another side to form a rectangle...2011-09-04
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    Is $(a+b)^2+(a-b)^2=2a^2+2b^2$ intuitive?2011-09-04
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    @joriki:I guess so,but I haven't fully understood how are you representing diagonals to $(a \pm b)^2$?!2011-09-04
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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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    http://en.wikipedia.org/wiki/Parallelogram_law2011-09-04
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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