By similarity of $ABC$, $AC_1B_1$, $BC_1A_1$,$CA_1B_1$, we have $$\prod_{cyc}\frac{B_1A_2}{A_2C_1}=$$$$=\prod_{cyc}\frac{(B_1C_1+B_1A-C_1A)/2}{(B_1C_1+AC_1-AB_1)/2}=$$$$=\prod_{cyc}\frac{BC+AB-AC}{BC+AC-AB}=1$$So by Ceva's theorem the three lines $A_1A_2,B_1B_2,C_1C_2$ concur.

When I first read the problem, I misread and thought we had to prove that $AA_2,BB_2,CC_2$ concur. Which happens to be true as well, by a similar argument using Ceva (or trigonometric Ceva).

Note that $AA_2$, $BB_2$, and $CC_2$ concur at the isogonal conjugate of the Gergonne point of $ABC$ so we're done by the converse of cevian nest.

Oops my solution same as above's one We show $AA_2,BB_2,CC_2$ concur. It's easy because $AA_2,BB_2,CC_2$ are isogonal ti the lines joining each vertex to the opposite intouch point. Hence $AA_2,BB_2,CC_2$ concur at the Nagel point of $ABC$. Using Cevian Nest on $ABC$ and $A_1B_1C_1$, we get $A_1A_2,B_1B_2,C_1C_2$ also concur