Geometry Problem involving Midpoints and Analytic Geoometry In triangle $ABC, AB = AC, D$ is the midpoint of $\overline{BC}$, E is the foot of the perpendicular from D to $\overline{AC}$, and F is the midpoint of $\overline{DE}$. Prove that $\overline{AF}$ is perpendicular to $\overline{BE}$.
I am required to solve this using analytic geometry and am recommended to place $A$ on the origin and $B$ on the point $(x,0)$, but don't know how to proceed from there.
 A: I've done it the opposite way. $A = (0,0)$ and $C = (1,0)$ (wlog I consider the length of $\overline {AC}$ to be $1$). Say the oriented angle from $\vec{AC}$ to $\vec{AB}$ be $\alpha$. Then since $\overline{AB} = \overline{AC} = 1$, $$B = (\operatorname{cos}\alpha, \operatorname{sin}\alpha)$$ $$D = \frac{1}{2}(C + B) = \frac{1}{2}(\operatorname{cos}\alpha + 1, \operatorname{sin}\alpha)$$ $E$ then is simply the projection of $D$ on the $Ox$, so $$E = (\frac{1}{2}(\operatorname{cos}\alpha + 1), 0) = \frac{1}{2}(\operatorname{cos}\alpha + 1, 0)$$ $$F = \frac{1}{2}(D + E) = \frac{1}{4}(2(\operatorname{cos} \alpha + 1), \operatorname{sin} \alpha)$$ Therefore $$\vec{AF} = \frac{1}{4}(2(\operatorname{cos \alpha + 1}), \operatorname{sin \alpha})$$ and $$\vec{EB} = B - E = \frac{1}{2}(\operatorname{cos} \alpha-1, 2 \operatorname{sin}\alpha)$$ Now when you calculate $\vec{AF}.\vec{EB}$ you get $\frac{1}{8}(2 \operatorname{cos}^2\alpha - 2 + 2\operatorname{sin}^2\alpha) = 0$, which is exactly what you want. 
A: Here's a start.
Set point $C$ at $(a,b)$ (with $b \neq 0$).  You can also specify $a \gt 0$ without loss of generality.
Now, you're in a position to define $D = ((a+x)/2, b/2).$
The slope of $\overline{AC}$ is $m_{AC} = b/a$, so the slope of a perpendicular to that line is $-a/b$.
Now, try to finish.
From there, you can determine the location of $E$, because you know $D$, and you know the slope of $\overline{DE}$.  (Write out an equation for the line $AC$, and for the line $DE$, and solve for $x,y$.)
Then, some more midpoint calculation, and finally show that the two calculated slopes are negative reciprocals of one another:
$$m_{AF} = -\frac{1}{m_{BE}}.$$
