Analytic Geometry proof of orthogonality in triangle geometry 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 $\overline{BE}$.

My first approach was to align the triangle in the first quadrant, on the x-coordinates and started calculating slopes and positions of points.  But then things got messy real fast, i'm afraid I'm approaching this problem the wrong way.  Is there a better way?  No trigonometry just yet!  Solutions are greatly appreciated.  Thanks in advance!

But maybe if you saw the geometric proof, you would change your mind about looking for an analytic proof of this fact. 
Take $K$ to be the midpoint of segment $CE$. Then $KF$ is a mid-segment in triangle $DCE$ and therefore $KF$ is parallel to $CD$. Furthermore, $CD$ is orthogonal to $AD$, which  means that line $KF$ is orthogonal to $AD$ too. Hence the lines $DE$ and $KF$ are altitudes in triangle $DKA$ and consequently point $F$ is the orthocenter of $DKA$. Therefore, the line $AF$ is the third altitude in triangle $DKA$ and so $AF$ is orthogonal to $DK$. The segment $DK$ however is mid-segment of triangle $BCE$ so $DK$ is parallel to $BE$. Thus $BE$ is orthogonal to $AF$.

