I am looking for a proof of the " begonia theorem". Let $D$, $E$, $F$ be points on respective (extended) sides $\overleftrightarrow{BC}$, $\overleftrightarrow{CA}$, $\overleftrightarrow{AB}$ of $\triangle ABC$, such that $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, $\overleftrightarrow{CF}$ are concurrent at a point $P$. Now, let $A^\prime$, $B^\prime$, $C^\prime$ be the reflections of $P$ over $\overleftrightarrow{EF}$, $\overleftrightarrow{FD}$, $\overleftrightarrow{DE}$, respectively. 

Prove that $\overleftrightarrow{AA^\prime}$, $\overleftrightarrow{BB^\prime}$, and $\overleftrightarrow{CC^\prime}$ are concurrent.

Please suggest a link or book or proof.

 A: (Extending comment as answer, as requested.)

This is GoGeometry's Dynamic Geometry Problem 974, which shows a nice animation. Linked there is a (zipped PostScript) note by Darij Grinberg that provides a proof of the Begonia Theorem using circle inversion. The proof is too long to reproduce, but I can give the steps ...
Grinberg first proves how an auxiliary point to a triangle leads to a construction of three circles through that point and another. (I've paraphrased slightly.)

Theorem 1. Given $\triangle ABC$ and point $P$, let $A^\prime \neq P$ be the "other" intersection of $\overleftrightarrow{AP}$ and $\bigcirc BCP$; similarly define $B^\prime$ and $C^\prime$. Let $X$, $Y$, $Z$ be the centers of $\bigcirc B^\prime C^\prime P$, $\bigcirc C^\prime A^\prime P$, $\bigcirc A^\prime B^\prime P$. Then, $\bigcirc PAX$, $\bigcirc PBY$, $\bigcirc PCZ$ are coaxal; that is, they have a common point different from $P$ (or else they touch at $P$).

Further,

Lemma 2. If $P^\prime$ and $Q^\prime$ are the respective images of $P$ and $Q$ in an inversion with center $O$, then the reflection of $O$ in $\overleftrightarrow{PQ}$ is the center of $\bigcirc P^\prime Q^\prime O$.

The note's Theorem 3 is the Begonia Theorem. For proof, Grinberg inverts with respect to $P$, tracking the images of $A$, $B$, $C$, $D$, $E$, $F$, $A^\prime$, $B^\prime$, $C^\prime$. (Here, I'm using the names of points as in the figure given in the question; Grinberg has slightly different notation.)

*

*By Lemma 2, $A^\prime$ inverts to the center ($X$) of the circular image of $\overline{EF}$; similarly, $B^\prime$, $C^\prime$ invert to centers ($Y$, $Z$) of the images of the other edges.


*By Theorem 1, $\bigcirc PAX$, $\bigcirc PBY$, $\bigcirc PCZ$ have "another" point in common; the inverse of that point is the desired Begonia point.

In his "Unpublished Notes" entry regarding the result, Grinberg references a proof by Jean-Pierre Ehrmann posted as message #8039 of the "Hyacinthos" Yahoo discussion group. (See also some follow-up discussion, at least through message #8049.) The one-line proof casually invokes pre-cevian (= trilinear polar?) and orthic triangles, harmonic quadruples, and isogonal conjugates. (Wow!)
