TITLE: Analytic Geometry proof of orthogonality in triangle geometry
QUESTION [5 upvotes]: 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!

REPLY [4 votes]: Here is a vectorial proof with a little trigonometry at the end. We can transform our objective thus:
$$\tag{2}AF \perp BE \ (a) \ \ \ \ \iff \ \ \ 2\vec{AF} \cdot
\vec{BE}=0 \ (b)$$
(symbol $\cdot$ meaning "dot product"). The LHS of (2b) can be transformed thus:
$$(\vec{AD}+\vec{AE}) \cdot (\vec{BD}+\vec{DE})=\vec{AD}\cdot\vec{DE} + \vec{AE} \cdot \vec{BD} $$
(taking into account the orthogonality of AD and BD, and AE and DE).
Let $\alpha=\widehat{BAC}=\widehat{DAC}$ and $h$ be the length of altitude $AD$.
It is easy to establish that:
$$\begin{cases} \vec{AD} \cdot \vec{DE}&=&- h \times h \sin(\alpha) & \ \ \text{(the minus sign is essential)}\\ \vec{AE} \cdot \vec{BD} &=& h \cos(\alpha) \times h \tan(\alpha) & \end{cases}$$
(where $\times$ is the ordinary multiplication of real numbers). 
Adding these two results gives $0$ ; we have thus obtained the RHS of (2).