Consider an isosceles $ABC$ triangle ($AB=AC$), such that if $D$ is the midpoint of base $BC$, then $BC=AD$. Let $E$ be a point on segment $AB$, such that $DE\perp AB$. Furthermore let $F$ be the midpoint of segment $DE$.
Prove that lines $AF, CE$ are perpendicular!
It is not that difficult to see that the slope of $CE$ is $-2/9$ and the slope of $AF$ is $9/2$. So we are done, but does anyone have a proof sithout calculus, a more Euclidean one, with angle chasing?
On
Not much of a proof but another convincing way to show that $\overline{AF}\bot\overline{CE}$:
Consider $Q$ as the midpoint of $\overline{AC}$, drawing a circle with radius $\overline{QA}$, it can be seen that points $C, A\ \& \ w$, where $w$ is the intersection between $FA$ and $CE$, all lie on the circle.
By virtue of Thales' theorem, we know definitely that $\overline{AF}\bot\overline{CE}$.
Let $G$ be the midpoint of $CD$. We get $AFG \sim ADC$ from $AEFD \sim ADGC$. So $\angle AFG = \pi/2$, and since $FG \parallel EC$ we have done.