We have $DE \perp BC$ and $DF \perp AB$ (because $\Gamma_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$).

Figure 4

Figure 5

Hence the quadrilateral $BEDF$ is cyclic and let $\Gamma_3$ be its circumcircle.
Moreover we have the equalities:
$$BE = BF \text{ and } DE = DF. \tag{1}$$

The $EF$ is the common chord of the circles $\Gamma_2$ and $\Gamma_3$ and $KG$ is the common chord of the circles $\Gamma_1$ and $\Gamma_2$. Since $EF$ and $KG$ meet at $M$, the common chord $BD$ of the circles $\Gamma_1$ and $\Gamma_3$ will pass through $M$ (radical center of the circles $\Gamma_1, \Gamma_2, \Gamma_3$).
Since the orthogonal triangles $AFD$, $CED$ are equal ($FD = ED$, $\angle FAD = \angle ECD$), we have $DA = DC$. Hence $DO$ is the perpendicular bisector of $AC$, and since $DO \perp KG$, it follows that $AC \parallel KG$ and $\angle ACD = \angle MND$.
Since $BD$ bisects the angle $\angle FBE$ and $BDCA$ is inscribed into the circle $\Gamma_1$ we have $\angle MBC = \angle FBM = \angle ACD = \angle MND$. Hence $BCNM$ is cyclic.

We have $DE \perp BC$ and $DF \perp AB$ (because $\Gamma_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$). Hence the quadrilateral $BEDF$ is cyclic and let $\Gamma_3$ be its circumcircle.
We consider the inversion with respect to the circle $\Gamma_2$ (with center $D$). Since circle $\Gamma_3$ is passing through $D$, its image is the line of common chord $FE$, and therefore:
The image of $B$ belongs to the line $FE$. (5)
Since circle $\Gamma_1$ is passing through $D$, its image is the line of common chord $KG$, and therefore:
The image of $B$ belongs to the line $KG$. (6)
From (5), (6) we conclude that the image of $B$ under the inversion we have considered is the point $M$. Also, $C$ has image the point $N$. Hence the quadrilateral $BMNC$ is cyclic.