Let $J$ be the First Isodynamic Point and let $A', B', C'$ be the reflections of $J$ over lines $BC,CA,AB$. Show that the lines $AA', BB', CC'$ concur at the First Fermat Point of triangle $ABC$.
My Progress:
Claim $1$: $A'$ lies on the $A-$ appolonion circle and similarly.
Proof: We know that $J$ lies on the $A-$ appolonion circle. So by definition, $\frac{JB}{JC}=\frac{AB}{AC}$. And since $A'$ is reflection of $J$, we have $\frac{A'B}{A'C}=\frac{AB}{AC}$
Claim $2$: It is well known that the pedal triangle of the first isodynamic point is equilateral. So if $LMN$ is the pedal triangle of $J$, where $L,M,N$ lie on $BC,CA,AB$ respectively, then it is sufficient to prove that $AA' \perp MN$ and similarly.
Proof: It is well known that the perpendiculars from the vertices to the sidelines of the pedal triangle of a point concur at the isogonal conjugate of the point. And Fermat Point and Isodynamic Point are isogonal conjugates.
Now I'm not sure how to prove that $AA' \perp MN$ or if Claim $1$ is even helpful. Please help me to proceed ahead. Thanks in advance.
Best Answer
Note that $AC'=AJ=AB'$ and as $J$ is the first isodynamic point, we get $\triangle LMN$ is equilateral and thus, considering a dilation centered at $J$ of factor $2$, we get $A'C'=A'B'$ and $B'C'\| MN$ and thus, $AA'$ is perpendicular bisector of $B'C'$ and so, $AA'\perp MN$. Hence, $\{AA'\cap BB'\cap CC'\}=F_1$ as $F_1$ and $J$ are isogonal conjugates. $$\tag*{$\blacksquare$}$$