Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $A B M, M N C$, and $N D A$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $C B$ is also tangent to $k_{3}$. ! Figure 1 To this end, let $X$ denote the point of intersection of $g$ and $h$. Then $A B C X$ and $A B C D$ are circumscribed quadrilaterals, whence
$$C D - C X = (A B + C D) - (A B + C X) = (B C + A D) - (B C + A X) = A D - A X$$
i.e.
$$A X + C D = C X + A D$$
which in turn reveals that the quadrilateral $A X C D$ is also circumscribed. Thus $h$ touches indeed the circle $k_{3}$. Moreover, we find that $\angle I_{3} C I_{1} = \angle I_{3} C X + \angle X C I_{1} = \frac{1}{2}(\angle D C X + \angle X C B) = \frac{1}{2} \angle D C B = \frac{1}{2}(180^{\circ} - \angle M C N) = 180^{\circ} - \angle M I_{2} N = \angle I_{3} I_{2} I_{1}$, from which we conclude that $C, I_{1}, I_{2}, I_{3}$ are concyclic. Let now $L_{1}$ and $L_{3}$ be the reflection points of $C$ with respect to the lines $I_{2} I_{3}$ and $I_{1} I_{2}$ respectively. Since $I_{1} I_{2}$ is the angle bisector of $\angle N M C$, it follows that $L_{3}$ lies on $g$. By analogous reasoning, $L_{1}$ lies on $g$. Let $H$ be the orthocenter of $\triangle I_{1} I_{2} I_{3}$. We have $\angle I_{2} L_{3} I_{1} = \angle I_{1} C I_{2} = \angle I_{1} I_{3} I_{2} = 180^{\circ} - \angle I_{1} H I_{2}$, which entails that the quadrilateral $I_{2} H I_{1} L_{3}$ is cyclic. Analogously, $I_{3} H L_{1} I_{2}$ is cyclic. Then, working with oriented angles modulo $180^{\circ}$, we have
$$\angle L_{3} H I_{2} = \angle L_{3} I_{1} I_{2} = \angle I_{2} I_{1} C = \angle I_{2} I_{3} C = \angle L_{1} I_{3} I_{2} = \angle L_{1} H I_{2},$$
whence $L_{1}, L_{3}$, and $H$ are collinear. By $L_{1} \neq L_{3}$, the claim follows.

Comment. The last part of the argument essentially reproves the following fact: The Simson line of a point $P$ lying on the circumcircle of a triangle $A B C$ with respect to that triangle bisects the line segment connecting $P$ with the orthocenter of $A B C$.