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.

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:
$$m_{AF} = -\frac{1}{m_{BE}}.$$
