We have a $\triangle ABC$ and a $\triangle A_1B_1C_1$. The segments $CL$ and $C_1L_1$ are angle bisectors. If $\triangle ALC \sim \triangle A_1L_1C_1$, I should show that $\triangle ABC \sim \triangle A_1B_1C_1$.
From the similarity, we have $\dfrac{AL}{A_1L_1}=\dfrac{CL}{C_1L_1}=\dfrac{AC}{A_1C_1}$. The only way I see from here is to show that $\triangle LBC \sim \triangle L_1B_1C_1$. Is this necessary for the solution?
If $\triangle ALC \sim \triangle A_1L_1C_1$ then $\angle A = \angle A_1$. Also, $\angle ACL = \angle A_1C_1L_1$. But $\angle C = 2\angle ACL $ and $\angle C_1 = 2\angle A_1C_1L_1 $, so $\angle C = \angle C_1$, etc.