Let ABC be a right triangle with A the right angle and let D be the mid-point of side AB. If E is the foot of the perpendicular from A to CD and F is the mid-point of CE, prove that BE is perpendicular to AF.

WLOG
let the corordinates of A,B,C be
A: (0,0)
B: (0,2)
C: (x,0)

then we obviously have
D: (0,1)

let E be at (p,q) then we have

p^2+q^2=x^2/(x^2+1)
q=1-(p/x)
solving we get
p=x/(x^2+1) q=x^2/(x^2+1)
thus
E: ( x/(x^2+1) , x^2/(x^2+1) )

then if F is at (p,q) we get
p=( x/(x^2+1) + x)/2
p=(x^3+2x)/(2*(x^2+1))
q=x^2/(2*(x^2+1))
thus we have
F: ( (x^3+2x)/(2*x^2+2) , x^2/(2*x^2+2) )

let slope of BE=m1 and slope of AF=m2

if m1*m2=-1 then they are perpindicular

m1=( x^2/(x^2+1) - 2 )/( x/(x^2+1) )
m1=( (x^2-2x^2-2)/(x^2+1) )*(x^2+1)/x
m1=-(x^2+2)/x

m2=( x^2/(2*x^2+2) )/( (x^3+2x)/(2*x^2+2) )
m2=(x^2/(x^3+2x)=x^2/(x*(x^2+2))=x/(x^2+2)

m1*m2= ( -(x^2+2)/x )*( x/(x^2+2) )
m1*m2=-1

thus BE and AF are perpendicular

Posted by Daniel on 2008-10-06 18:01:34