$AX \parallel BZ$ implies $\frac{AN}{NB} = \frac{AX}{BZ} = \frac{CX}{CB}$, and $AX \parallel CY$ implies $\frac{AM}{MC} = \frac{AX}{CY} = \frac{BX}{BC}$. Hence,
$$\frac{AN}{NB} + \frac{AM}{MC} = \frac{CX}{CB} + \frac{BX}{BC} = 1. \quad (1)$$
Now let $a$ denote the sidelength of $ABC$. Using (1), one finds
$$\frac{a^2}{NB \cdot MC} = \left(1 + \frac{AN}{NB}\right) \left(1 + \frac{AM}{MC}\right) = 2 + \frac{AN \cdot AM}{NB \cdot MC},$$
Therefore,
$$AN \cdot AM = a^2 - 2 \cdot NB \cdot MC. \quad (2)$$
Now the Law of Cosines implies $MN^2 = AM^2 + AN^2 - AM \cdot AN$ which together with (2) gives
$$\begin{aligned} MN^2 &= AM^2 + AN^2 - a^2 + 2 \cdot NB \cdot MC \\ &= (a - NB)^2 + (a - MC)^2 - a^2 + 2 \cdot NB \cdot MC \\ &= (NB + MC - a)^2. \end{aligned}$$
Hence, $MN = NB + MC - a$ (it can be easily seen by (2) that $NB + MC - a > 0$). Finally $MN + BC = NB + MC$ implies that the quadrilateral $MNBC$ has an incircle, thus the incircle of $ABC$ is tangent to the segment $MN$. Done.