Geometry, two perpendicular lines

"Let $\hat{ABC}$ be an isosceles triangle with $AB=AC$. $D$ is a point on $BC$ such that $DC=DB$ (middle of $BC$). $E$ is the projection of $D$ on $AC$ and $F$ the middle of $DE$. Prove, using vectors that $AF$ and $BE$ are perpendicular."

The start should be that $AF$ and $BE$ are perpendiculars $\iff$ the scalar product of $AF$ and $BE = 0$. Any nice solution using vector geometry? A synthetic one or with complex numbers would be appreciated as well!

• "DA=DB" does not mean "middle of BC" – sds Jun 11 '14 at 16:53
• It was a mistake! It must be DB=DC. – user146371 Jun 11 '14 at 16:58

Let $A$ be your origin, $\vec b=AB$, $\vec c=AC$. Then
1. Express the isosceles condition in terms of $\vec b$ and $\vec c$.
2. Express vector $AD$ in terms of $\vec b$ and $\vec c$.
3. Likewise for vector $AE$
4. Likewise for vector $AF$
5. Likewise for vector $BE$
Let's express almost all the vectors as relative to point $D$: $$\begin{array}{ccccc} \vec{a} \equiv \vec{DA} & & \vec{c} \equiv \vec{DC} & & \vec{b} \equiv \vec{DC} = -\vec{c}\\ \vec{e} \equiv \vec{DE} & & \vec{f} \equiv \vec{DF} = \frac{1}{2} \vec{e} & &\vec{g} \equiv \vec{AC} = \vec{c}\vec{a} \end{array}$$ Then since $\triangle BAC$ is isosceles, $AD$ is an altitude, hence $$\begin{array}{c} \vec{c} \cdot \vec{a} = 0 \\ \vec{c} \cdot \vec{g} = \vec{c} \cdot \vec{g} = \vec{c} \cdot \vec{c} - \vec{c} \cdot \vec{a} = |c|^2 \end{array}$$ Point $E$ is on line $AC$ so for some real $\kappa$ $$\vec{e} = \vec{c} + \kappa (\vec{c} - \vec{a}) = \vec{c} + \kappa \vec{g}$$ We now enforce the fact that $DE \perp AC$ by $$\begin{array}{l} \vec{e} \cdot \vec{g} = 0 \\ (\vec{c} + \kappa \vec{g}) \cdot \vec{g} = 0 \\ \vec{c} \cdot \vec{g} + \kappa |g|^2 = 0 \\ \kappa = -\frac{\vec{c} \cdot \vec{g}}{g^2} = -\frac{c^2}{g^2}\\ \vec{e} = \vec{c} - \frac{c^2}{g^2} \vec{g} \end{array}$$ And $\vec{f} = \frac{1}{2}\vec{e}$. Now we express $AF$ and $BE$ $$\begin{array}{l} \vec{AF} = \vec{f} - \vec{a} = \vec{f} - \vec{c} + \vec{g} \\ 2 g^2 \vec{AF} = g^2 \vec{c} - c^2 \vec{g} - 2 g^2 \vec{c} + 2 g^2 \vec{g} \\ 2 g^2 \vec{AF} = - g^2 \vec{c} - c^2 \vec{g} + g^2 \vec{c} \\ \vec{BE} = \vec{e} - \vec{b} = \vec{e} + \vec{c} \\ g^2 \vec{BE} = 2 g^2 \vec{c} - c^2 \vec{g} \end{array}$$ Finally, we examine the dot product. Since $|g|$ is the positive length of $AC$, if the dot product of $2 g^2 \vec{AF}$ and $g^2 \vec{BE}$ is zero, that will prove that $AF \perp BE$. $$\begin{array}{l} 4 g^4 \vec{BE} \cdot \vec{AF} = \left( 2 g^2 \vec{c} - c^2 \vec{g} \right) \cdot \left( 2 g^2 \vec{g} - g^2 \vec{c} - c^2 \vec{g} \right) \\ = 4 g^4 \vec{c} \cdot \vec{g} -2 g^4 c^2 - 2 g^2 c^2 \vec{c} \cdot \vec{g} - 2 g^4 c^2 + c^2 g^2 \vec{c} \cdot \vec{g} + g^2 c^4 \\ = 4 g^4 c^2 -2 g^4 c^2 - 2 g^2 c^4 - 2 g^4 c^2 + c^2 g^2 c^2 + g^2 c^4 =0 \end{array}$$