TITLE: Justify this is locus is indeed a parabola
QUESTION [5 upvotes]: This problem consists of a geometric construction and proving that the locus generated is a parabola. Here's the construction:


Trace a circumference with radius $r$ and center $A$.

From any point $C$ on $\bigcirc A$, trace the line $\overleftrightarrow{AC}$. Let $D$ be a point on this line such that $C$ is the midpoint of $\overline{AD}$.

Let $E$ be a point on $\bigcirc A$. Trace the line $\overleftrightarrow{ED}$.

Let $E^\prime$ be reflection of $E$ in $\overleftrightarrow{AC}$. Trace the line $\overleftrightarrow{AE^\prime}$.

Let $F$ the intersection of $\overleftrightarrow{AE^\prime}$ with $\overleftrightarrow{ED}$.


Prove the locus described by $F$, as the point $E$ moves through the circumference, is a parabola.


It might seem simple, but this problem has been giving me some troubles trying to find the key idea.

REPLY [4 votes]: Let $G$ be the foot of the perpendicular from $F$ to the perpendicular bisector of $\overline{AD}$. When $\overleftrightarrow{E^\prime A}$ is not parallel to that bisector (that is, for $\overleftrightarrow{E^\prime A} \not\perp \overline{AC}$), let $P$ be the point where the two lines meet. 

Then,
$$
\frac{|\overline{AF}|}{|\overline{AE}|}
\;\underbrace{\phantom{\huge|}\quad=\quad\phantom{\huge|}}_{\triangle FAE\sim\triangle FPD}\;
\frac{|\overline{PF}|}{|\overline{PD}|} 
\;\underbrace{\phantom{\huge|}\quad=\quad\phantom{\huge|}}_{\triangle APD \text{ is isos.}}\;
\frac{|\overline{PF}|}{|\overline{PA}|} 
\;\underbrace{\phantom{\huge|}\quad=\quad\phantom{\huge|}}_{\triangle PFG \sim \triangle PAC} \;
\frac{|\overline{FG}|}{|\overline{AC}|}$$
But,
$$\overline{AE} \cong \overline{AC}$$
So,
$$\overline{AF} \cong \overline{FG}$$
Thus, 

$F$ is equidistant from $A$ and the perpendicular bisector of $\overline{AD}$. (Note that this is trivially true when $\overleftrightarrow{E^\prime A}\perp\overline{AD}$, as well, since then $F$ and $E$ coincide as a vertex of square $\square ACGF$.) Consequently, its locus, by definition, is the parabola with the latter elements as focus and directrix. $\square$