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$|DF|=|FB|$

$|AE|=|EC|$

What is $|AC|^2 + |BD|^2= \ ?$

I asked one of my classmates for the solution, he said:

$$|AC|^2+|BD|^2+4 \cdot3^2 = 7^2 +4^2 +5^2+6^2$$

He didn't tell me how he obtained it, either because he wouldn't want to tell me (I remotely know him) or he just memorized the formula.

Provided that this is a high school geometry problem, can you show me how above equation is obtained in simple terms?

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There is identity which states that for every quadrilateral $ABCD$ we have: $$AB^2+BC^2+CD^2+DA^2 = AC^2+BD^2+4EF^2$$


Proof: Remember that $XY^2 = \vec{XY}^2 = (Y-X)^2 = Y^2-2X\cdot Y+X^2$

Also remember also that $E = {1\over 2}(A+C)$ and $F = {1\over 2}(B+D)$.

So the left side is $$2A^2+2B^2+2C^2+2D^2-2A\cdot B-2B\cdot C-2C\cdot D-2D\cdot A$$

and the right side is $$A^2+C^2-2A\cdot C +B^2+D^2-2B\cdot D +(B+D-A-C)^2$$

Now it should not be difficult to see they are the same...


So use it and you are done.

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