Let $\angle BAC = 2\alpha$, $\angle ABC = 2\beta$, $\angle BCA = 2\gamma$. Since $AM = AN$, we obtain $\angle AMN = \angle ANM$. We have $\angle AMN = \angle ACN$ and $\angle MNC = \angle MAC$ (as inscribed angles). Further,
$$\begin{aligned} \angle AKN &= \angle MAK + \angle AMK = \\ &= \angle MAC + \angle AMN = \\ &= \angle MNC + \angle ANM = \angle ANC. \end{aligned}$$
It follows that the triangles $AKN$ and $ANC$ are similar. Similarly, the triangles $ALM$ and $AMB$ are similar. Then $\frac{AN}{AC} = \frac{AK}{AN}$, so $AI^2 = AN^2 = AK \cdot AC$.
In the same manner we see that $AI^2 = AL \cdot AB$, so $AK \cdot AC = AL \cdot AB$, which shows that $L$, $K$, $C$, $B$ are concyclic. Then $\angle AKL = \angle ABC = 2\beta$. Moreover, since $AI : AK = AC : AI$, it follows that the triangles $AIK$ and $ACI$ are similar. Then $\angle AIK = \angle ICA = 0.5\angle BCA = \gamma$. Since $\angle IKC = \angle IAK + \angle AIK = \alpha + \gamma$, we have
$$\begin{aligned} \angle IKL &= 180^\circ - \angle AKL - \angle IKC = 180^\circ - 2\beta - \alpha - \gamma = \\ &= 180^\circ - 2(\alpha + \beta + \gamma) + \alpha + \gamma = [\alpha + \beta + \gamma = 90^\circ] = \alpha + \gamma = \angle IKC. \end{aligned}$$
Similarly, $\angle KLI = \angle ILB$. Therefore, all bisectors of the quadrilateral $LBCK$ intersect at the same point $I$. Hence $LBCK$ is a circumscribed quadrilateral. Thus $MN$ touches the inscribed circle of the triangle $ABC$.