Solution

Here is a proof by Bui Quang Tuan:

Suppose (O') is circumcircle of triangle QM₄N₄.

Let Y and Z be the intersections with MN (L₁) of lines M₁M₂ (L_m) and N₁N₂ (L_n), respectively. Denote the circumcircles of MXQ and NXQ as (O_y), (O_z) respectively.

Y is the intersection of M₁M₂ and MX. Therefore Y is radical center of (O), (O_m) and (O_y). This means that

Y is the intersection of QM₄ with MX. Therefore Y is radical center of (O'), (O_m) and (O_y) implying that

The radical center of (O), (O') and (O_m) is the intersection of the common chord of (O), (O_m) and the common chord of (O') and (O_m). By (1) and (2) it is point Y. So Y is the radical center of (O), (O') and (O_m) meaning that the common chord of (O), (O') passes through Y.

Similarly, we can show that common chord of (O), (O') passes through Z. Since both Y and Z are on L₁, L₁ is the common chord of (O), (O'). Therefore M₃, N₃ are the intersections of the two circles (O), (O'). This exactly means that Q, M₃, N₃, M₄, N₄ are concyclic on (O').