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!


We may embed the construction in the complex plane, assuming $D=0$, $C=1, B=-1$ and $A=ri$. With such assumptions we may compute $E$, then $F=\frac{E}{2}$ and check that $\frac{F-A}{E-B}$ is a purely imaginary number, proving $AF\perp BE$.
