Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $ABM$, $MNC$, and $NDA$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $CB$ is also tangent to $k_{3}$.

Figure 1

To this end, let $X$ denote the point of intersection of $g$ and $h$. Then $ABCX$ and $ABCD$ are circumscribed quadrilaterals, whence
$$CD - CX = (AB + CD) - (AB + CX) = (BC + AD) - (BC + AX) = AD - AX,$$
i.e.
$$AX + CD = CX + AD$$
which in turn reveals that the quadrilateral $AXCD$ is also circumscribed. Thus $h$ touches indeed the circle $k_{3}$.

Moreover, we find that $\angle I_{3} C I_{1} = \angle I_{3} CX + \angle XCI_{1} = \frac{1}{2}(\angle DCX + \angle XCB) = \frac{1}{2} \angle DCB = \frac{1}{2}(180^{\circ} - \angle MCN) = 180^{\circ} - \angle MI_{2}N = \angle I_{3} I_{2} I_{1}$, 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 $\angle NMC$, 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.

We start by proving that $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic.

Figure 2

To this end, notice first that $I_{2}, M, I_{1}$ are collinear, as are $N, I_{2}, I_{3}$ (see Figure 2). Denote by $\alpha, \beta, \gamma, \delta$ the internal angles of $ABCD$. By considerations in triangle $CMN$, it follows that $\angle I_{3} I_{2} I_{1} = \frac{\gamma}{2}$. We will show that $\angle I_{3} C I_{1} = \frac{\gamma}{2}$, too. Denote by $I$ the incenter of $ABCD$. Clearly, $I_{1} \in BI$, $I_{3} \in DI$, $\angle I_{1} A I_{3} = \frac{\alpha}{2}$.

Using the abbreviation $[X, YZ]$ for the distance from point $X$ to the line $YZ$, we have because of $\angle BAI_{1} = \angle IAI_{3}$ and $\angle I_{1} AI = \angle I_{3} AD$ that
$$\frac{[I_{1}, AB]}{[I_{1}, AI]} = \frac{[I_{3}, AI]}{[I_{3}, AD]}$$
Furthermore, consideration of the angle sums in $AIB$, $BIC$, $CID$ and $DIA$ implies $\angle AIB + \angle CID = \angle BIC + \angle DIA = 180^{\circ}$, from which we see
$$\frac{[I_{1}, AI]}{[I_{3}, CI]} = \frac{I_{1}I}{I_{3}I} = \frac{[I_{1}, CI]}{[I_{3}, AI]}$$
Because of $[I_{1}, AB] = [I_{1}, BC]$, $[I_{3}, AD] = [I_{3}, CD]$, multiplication yields
$$\frac{[I_{1}, BC]}{[I_{3}, CI]} = \frac{[I_{1}, CI]}{[I_{3}, CD]}$$
By $\angle DCI = \angle ICB = \gamma / 2$ it follows that $\angle I_{1} CB = \angle I_{3} CI$ which concludes the proof of the above statement.

Let the perpendicular from $I_{1}$ on $I_{2} I_{3}$ intersect $g$ at $Z$. Then $\angle MI_{1}Z = 90^{\circ} - \angle I_{3} I_{2} I_{1} = 90^{\circ} - \gamma / 2 = \angle MCI_{2}$. Since we have also $\angle ZMI_{1} = \angle I_{2} MC$, triangles $MZI_{1}$ and $MI_{2}C$ are similar. From this one easily proves that also $MI_{2}Z$ and $MCI_{1}$ are similar. Because $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic, $\angle MZI_{2} = \angle MI_{1}C = \angle NI_{3}C$, thus $NI_{2}Z$ and $NCI_{3}$ are similar, hence $NCI_{2}$ and $NI_{3}Z$ are similar. We conclude $\angle ZI_{3}I_{2} = \angle I_{2}CN = 90^{\circ} - \gamma / 2$, hence $I_{1} I_{2} \perp ZI_{3}$. This completes the proof.