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Squaring sides (Posted on 2005-05-04) Difficulty: 4 of 5
Take any convex quadrilateral ABCD, with diagonals AC and BD. Pick E so ABCE is a parallelogram. Prove that AB+BC+CD+DA= AC+BD+DE.

Given the same quadrilateral, let P and Q be the midpoints of AC and BD. Now prove that AB+BC+CD+DA= AC+BD+4PQ.

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 AB+BC+CD+DA=
[xa + ya] + [xc] + [(xd-xc) + yd] + [(xd-xa) + (yd-ya)]=
2xa + 2xc + 2xd - 2xc xd - 2xa xd + 2ya + 2yd - 2ya yd

But also AC + BD + DE=
[(xa-xc) + ya] + [xd + yd] + [(xd-xa-xc) + (yd-ya)]=
2xa + 2xc + 2xd - 2xc xd - 2xa xd + 2ya + 2yd - 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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