Let ABC be a triangle and D the foot of the altitude from A. Let E and F lie on a line passing through D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D. Let M and N be the midpoints of the segments BC and EF, respectively. Prove that AN is perpendicular to NM.

I keep accidentally deleting before posting.  Here's the extremely short version (all algebra steps omitted.)
A=(0,a)
B=(b,0)
C=(c,0)
M=((b+c)/2,0)
Points D,E,F are all on line y=mx
E and F are also points on circles. 
N=((b+c+2am)/(2(1+m^2)),m(b+c+2am)/(2(1+m^2)))

Slope AN = -1(2a-mb-mc)/(b+c+2am)
Slope MN = (b+c+2am)/(2a-bm-cm)

If anyone would like me to fill in details later I will.

Posted by Jer on 2012-12-26 23:49:36