Simple Angle-Chasing gives us $$\angle BKL=\angle BMN=\angle BMC+\angle CMN=\angle BDC+\angle DBN=\angle BLC$$Hence, $DC$ is tangent to $(AKL)$. Similarly, $DA$ is tangent to $(AKL)$ as well. Now, look at the homothety which takes $K$ and $L$ to $M$ and $N$, respectively. It takes $AD$ to the tangent at $M$ to $k$, say $\ell_M$, and $CD$ to the tangent at $N$ to $k$, say $\ell_N$. Since $AD\cap CD=D$, we find that $\ell_M\cap \ell_N$ lies on $BD$. This means $BD$ is symmedian in the triangle $BMN$. Yields, $(MON)$ bisects the segment $BD$.