Solution:

Let $A = 2\alpha$. Then $\angle A_1AC = \angle BAA_1 = \alpha$. Thus
$$\angle A_1B_1C = \alpha = \angle BB_1A_1 = \angle A_1C_1C = \angle BC_1A_1$$
We also have $\angle B_1CQ = \angle AA_1B_1 = \beta$, say. It follows that triangles $MA_1B_1$ and $QCB_1$ are similar and hence
$$\frac{QC}{MA_1} = \frac{B_1C}{B_1A_1}$$

Similarly, triangles $ACM$ and $C_1A_1M$ are similar and we get
$$\frac{AC}{AM} = \frac{C_1A_1}{C_1M}$$
Using the point $P$, we get similar ratios:
$$\frac{PB}{MA_1} = \frac{C_1B}{A_1C_1}, \quad \frac{AB}{AM} = \frac{A_1B_1}{MB_1}$$
Thus,
$$\frac{QC}{PB} = \frac{A_1C_1 \cdot B_1C}{C_1B \cdot B_1A_1}$$
and
$$\begin{aligned} \frac{AC}{AB} & = \frac{MB_1 \cdot C_1A_1}{A_1B_1 \cdot C_1M} \\ & = \frac{MB_1}{C_1M} \frac{C_1A_1}{A_1B_1} = \frac{MB_1}{C_1M} \frac{C_1B \cdot QC}{PB \cdot B_1C} \end{aligned}$$
However, triangles $C_1BM$ and $B_1CM$ are similar, which gives
$$\frac{B_1C}{C_1B} = \frac{MB_1}{MC_1}$$
Putting this in the last expression, we get
$$\frac{AC}{AB} = \frac{QC}{PB}$$
We conclude that $PQ$ is parallel to $BC$.