Draw through $C$, a line $l$ parallel to $AB$.Extend $C_1A_1$ to meet line $l$ at some point $K$,and $C_1B_1$ at some point $L$.Now, $\angle BC_1A_1=\angle A_1KC$ and $\angle AC_1B_1=\angle CB_1L$ by parallelity.This implies that $\angle A_1KC=\angle CB_1L$ and thus it is sufficient to prove that $KC=CL$. Note that by parallelity, $KC=\frac{C_1B.A_1C}{BA_1}$ and similarly $CL=\frac{AC_1.B_1C}{AB_1}$. Thus, it is sufficient to prove that $AB_1.A_1C.C_1B=BA_1.AC_1.B_1C$,which is trivial by Ceva's Theorem.$\Square$