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.
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 xaxis 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] + [(xdxc)² + yd²] + [(xdxa)² + (ydya)²]=
2xa² + 2xc² + 2xd²  2xc xd  2xa xd + 2ya² + 2yd²  2ya yd
But also AC² + BD² + DE²=
[(xaxc)² + ya²] + [xd² + yd²] + [(xdxaxc)² + (ydya)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 20050505 17:42:01 