Lemma: If triangle $XYZ$ has circumcenter $O$ and orthocenter $H$, then $h=x+y+z-2o$. Proof: Use vectors. As $\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$ We have $\overrightarrow{XA}+\overrightarrow{XB}+\overrightarrow{XC}=3\overrightarrow{XO}+\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=3\overrightarrow{XO}+\overrightarrow{OH}$. So $\overrightarrow{XH}=\overrightarrow{XO}+\overrightarrow{OH}=\overrightarrow{XA}+\overrightarrow{XB}+\overrightarrow{XC}-2\overrightarrow{XO}$

rename $A_1, B_1, C_1$ to $D, E, F$ and let the incenter be $I$. use complex numbers with incircle as the unit circle. we have $a=\frac{2ef}{e+f}, b=\frac{2fd}{f+d}, c=\frac{2de}{d+e}$ Let the circumcenters of $EAF, FBD, DCE$ be $O_1, O_2, O_3$ respectively As $AEIF$ is cyclic with circumcenter being the midpoint of $AI$, we have $o_1=\frac{ef}{e+f}$ With the lemma we have $h_1=a+e+f-2o_1=e+f$ for $o_1=\frac{a+0}{2}$ and similarly we have $h_2=f+d, h_3=d+e$. Therefore $H_1H_2=|(e+f)-(f+d)|=|e-d|=DE$, similarly $H_2H_3=EF, H_3H_1=FD$.

Let $I$ be the incentre of triangle $ABC$. Then $H_1, H_2, H_3$ are the reflections of I over $B_1C_1, A_1C_1, A_1B_1$ respectively. From there, we see that $H_1H_2H_3$ is thus a dilation of the medial triangle of $A_1B_1C_1$ by a factor of 2 with centre $I$, and so it is congruent to triangle $A_1B_1C_1$.