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 Squaring sides (Posted on 2005-05-04)
Take any convex quadrilateral ABCD, with diagonals AC and BD. Pick E so ABCE is a parallelogram. Prove that AB²+BC²+CD²+DA²= AC²+BD²+DE².

Given the same quadrilateral, let P and Q be the midpoints of AC and BD. Now prove that AB²+BC²+CD²+DA²= AC²+BD²+4PQ².

 See The Solution Submitted by Federico Kereki Rating: 4.0000 (2 votes)

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 The analytic geometry way (part 1 only) | Comment 3 of 4 |

I'm sure there's a simpler way using geometry, but this way works:

Orient the quadrilateral so that B is at the origin and C is on the x-axis and name the points as follows:

A=(xa,ya) B=(0,0) C=(xc,0) D=(xd,yd)
Then E=(xa+xc,ya)

By pythagorean theorem AB²+BC²+CD²+DA²=
[xa² + ya²] + [xc²] + [(xd-xc)² + yd²] + [(xd-xa)² + (yd-ya)²]=
2xa² + 2xc² + 2xd² - 2xc xd - 2xa xd + 2ya² + 2yd² - 2ya yd

But also AC² + BD² + DE²=
[(xa-xc)² + ya²] + [xd² + yd²] + [(xd-xa-xc)² + (yd-ya)²]=
2xa² + 2xc² + 2xd² - 2xc xd - 2xa xd + 2ya² + 2yd² - 2ya yd

so they are equal.

I'm sure part 2 can be similarly solved, but it isn't worth doing out this way again.  I'd really like to see a nice purely geometric proof, though.  These are interesting results.

 Posted by Jer on 2005-05-05 17:42:01

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