<Tri-2024-2152>, hint

Let <BCM=x, MB=c, BC=a MX=y, MC=m

By angle chasing <XMA=<MCX=x 

AB/BC=AX/MC=c/a=y/m---> XA/MA=y/c=m/a. Hence Triangles XAC and XAM are 

similar. Hence <XAC=XAM=z.  MC is tangent to circle MXA. w

Let line NX cut w at P. Since MXAP are cyclic <APX=z=<XAM

Let line AC cut w at Q. Let M' be the midpoint of BC, Triangles MBM' and ABC are similar

<XNM=<MM'B=<BCA=<NCQ+<CQN...> <CQN=z=<APX  Hence P=Q.