Nice proof Heureka, my first attempt was by vectors but I couldn't find one, I intended to get back to it :) , but I switched my attention to a co-ordinate geometry approach.
I have two such solutions. The first had its origin at D with DA as the vertical axis. That worked but the algebra was messy in places. Much better it turns out is to have A as the origin with AC along the x-axis. (Diagram below).
Let B and C have co-ordinates (p, q) and (c, 0) respectively, then D will have co-ordinates
((p + c)/2, q/2).
From there, E will be ((p + c)/2, 0) and F ((p + c)/2, q/4).
The slope of AF will be \(\displaystyle \frac{q}{4}.\frac{2}{p+c} = \frac{q}{2(p+c)}\)
and the slope of BE\(\displaystyle\frac{q-0}{p-(p+c)/2}=\frac{2q}{p-c}\).
The product of those is
\(\displaystyle\frac{q}{2(p+c)}.\frac{2q}{(p-c)}=\frac{q^{2}}{p^{2}-c^{2}}\),
which, since
\(p^{2}+q^{2}=c^{2}\text{ so that } p^{2}-c^{2}=-q^{2}\)
is equal to -1.
Hence, AF and BE are at right angles to each other.
- Bertie
