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.

Say vertex A is at 0,0. Say B is at (Bx, By), C is at (Cx, Cy) and D is at (Dx, Dy).

Since E is the other corner of a parallelogram:
Ex = Bx + Cx
Ey = By + Cy

And since P and Q are midpoints:
Px = (Bx + Cx)/2 = Ex/2
Py = (By + Cy)/2 = Ey/2

Qx = (Ax + Dx)/2 = Dx/2
Qy = (Ay + Dy)/2 = Dy/2

The second proof is just a fancy way of saying "Prove that 4PQ^2 = DE^2" IF you assume that the first equation is provable/true (which I haven't done yet).

Well,
DE^2 = (Ex-Dx)^2 + (Ey-Dy)^2

PQ^2 = (Px-Qx)^2 + (Py-Qy)^2
PQ^2 = (Ex/2 – Dx/2)^2 + (Ey/2 – Dy/2)^2
PQ^2 = 1/4(Ex-Dx)^2 + 1/4(Ey-Dy)^2
4PQ^2 = (Ex-Dx)^2 + (Ey-Dy)^2
4PQ^2 = DE^2

Tada!

Posted by nikki on 2005-05-04 19:54:38