Let $S$ be the orthocenter of $\triangle A_1BC_1$. Let $N$ be the midpoint of $A_1C_1$. It's well-known that $H,N,S$ are collinear (since H is the antipode of A in A1BC1). Now just note $HS || B_1K$ since $\triangle A_1BC_1 \sim \triangle ABC$, and to finish we have $A_1C_1$ is antiparallel to $AC$ so the symmedian is the median hence $N$ lies on the symmedian of $ABC$, and we're done by thales or similar triangles.

Notice two facts: 1) $B-$symmedian in $ABC$ is the same as $B-$median in $A_1C_1$ since $AC$ and $A_1C_1$ are anti-parallel w/ respect to $\angle ABC$. 2) $B$ is $B_1-$excenter $\omega_{B_1}$ in $\Delta A_1B_1C_1$ so $K$ is tangency point. Under homothety centered at $K$ and radius 2 a midpoint of $KB_1$ maps to $B_1$. A midpoint of $A_1C_1$ maps to the tangency point of incircle $\omega$ of triangle $A_1B_1C_1$ and side $A_1C_1$. Finally, $B$ maps to the point opposite to $K$ in $\omega_{B_1}$. But three these points are collinear which clearly follows from another homothety centered at $B_1$ which maps $\omega$ to $\omega_{B_1}$.