Let $\pi_0$ denote the plane $ABC$, and for any points $X,Y$ on the plane $\pi_0$, let $\pi_{XY}$ denote the plane passing through points $X,Y$ and perpendicular to $\pi_0$. Consider only a single half-space on one side of $\pi_0$. Let $M_A$ and $N_A$ be the midpoints of $BC$ and $B_1C_1$ respectively; define $M_B,N_B,M_C,N_C$ similarly. Let $A'$ be the unique point on $\pi_{B_1C_1}$ in the half-space such that $B_1C_1A'$ is equilateral. This may be achieved by selecting $A'$ on the line through $N_A$ perpendicular to $\pi_0$ such that $A'N_A = \frac{\sqrt{3}}{2} B_1C_1$. Define $B',C'$ similarly. Note that $N_A$ (similarly $N_B,N_C$) is the projection of $A'$ (similarly $B',C'$) onto $\pi_0$. Claim 1: Rays $AA'$ and $BB'$ intersect. Proof: Note $A,B$ lie on both $\pi_0$ and $\pi_{AB}$. It suffices to show $$\frac{d(A', \pi_{AB})}{d(A', \pi_0)}=\frac{d(B', \pi_{AB})}{d(B', \pi_0)}$$Note that $$\frac{d(A', \pi_{AB})}{d(A', \pi_0)}=\frac{d(N_A, AB)}{A'N_A}=\frac{d(N_A,AB)}{\sqrt{3}\cdot C_1N_A} = \frac{\cos{\angle AC_1 B_1}}{\sqrt{3}}= \frac{\cos{\angle ACB}}{\sqrt{3}}$$and similarly for the RHS; the claim follows. $\square$ Now projecting rays $AA'$ and $BB'$ onto $\pi_0$, the intersection point is projected to $AN_A\cap BN_B$. Claim 2: $AN_A$ is the $A$-symmedian line of triangle $ABC$. Proof: Note that triangles $ABC$ and $AB_1C_1$ are similar (and related by homothety about $A$ and reflection about the angle bisector of $\angle BAC$); so the medians $AM_A$ and $AN_A$ are also related by reflection about the angle bisector of $\angle BAC$. Therefore, $AN_A$ is the $A$-symmmedian line. $\square$ Therefore, $AN_A$, $BN_B$, and $CN_C$ intersect at point $L$ which is the projection of all three intersection points $AA'\cap BB'$, $BB'\cap CC'$, and $CC' \cap AA'$. Thus the lines $AA', BB', CC'$ together with the line through $L$ perpendicular to $\pi_0$ are all concurrent at a single point.