Let $(ABC) \cap (A_2B_1C) = C \cup X$ I will show $X$ is the desired concurrency point. Note that $A_1B_1,B_1C_1,C_1A_1$ are the perpendicular bisectors of $\overline{CC_1},\overline{AA_1},\overline{BB_1}$ respectively. $$\measuredangle CXA_2 \equiv \measuredangle CB_1A_2 \equiv \measuredangle CB_1C_1 \equiv \measuredangle CBC_1 \equiv \measuredangle CBA \equiv \measuredangle CXA (mod 180^{\circ}) \implies X,A,A_2 \text{ are collinear}.$$$$\measuredangle ABX \equiv \measuredangle ACX \equiv \measuredangle B_1CX \equiv \measuredangle B_1A_2X \equiv \measuredangle C_1A_2A \equiv \measuredangle A_1A_2C \equiv \measuredangle B_2B_1C_1 \equiv \measuredangle C_1BB_2 \equiv \measuredangle ABB_2 (mod 180^{\circ}) \implies B,B_2,X \text{are collinear}.$$$$\measuredangle ACX \equiv \measuredangle B_1CX \equiv \measuredangle B_1A_2X \equiv \measuredangle C_1A_2A \equiv \measuredangle A_1A_2C_1 \equiv \measuredangle C_2C_1A_2 \equiv \measuredangle C_2C_1B_1 \equiv \measuredangle B_1CC_2 \equiv \measuredangle ACC_2 (mod 180^{\circ}) \implies C,X,C_2 \text{are collinear}$$.