If $(K_1)$ touches $AD$ at $N,$ we know that $MN$ passes through the C-excenter $E.$ Since lines $DK_1$ and $ AE$ are parallel, it follows that $ME \perp AE.$ Assume that $\triangle AMC$ is isosceles with apex $C.$ Then $\angle AEB=\angle AMC=90^{\circ}-\frac{_1}{^2}\angle C$ $\Longrightarrow$ quadrilateral $AEMB$ is cyclic $\Longrightarrow$ $\angle ABC=90^{\circ}.$ Therefore $\angle MAC=90^{\circ}-\frac{_1}{^2}\angle C-(90^{\circ}-\angle C)=\frac{_1}{^2}\angle C$ $\Longrightarrow$ $AM$ bisects $\angle BAD.$ Conversely, assume that $AM$ bisects $\angle BAD.$ Since $\angle MEB=\frac{_1}{^2}\angle MAB=\frac{_1}{^2}\angle C,$ it follows that quadrilateral $AEMB$ is cyclic, thus $\angle AEB=\angle AMC=90^{\circ}-\frac{_1}{^2}\angle C$ $\Longrightarrow$ $\triangle AMC$ is isosceles with apex $C.$

We will invert the diagram about $A$. For any object $x$, let $x'$ denote its image. Note that $K = (ABC)$ maps to a line $K' = \overline{B'C'}$ parallel to $\overline{AD'}$. Furthermore, $A,C',B',D'$ lie on the image of line $\overline{BC}$; thus $AC'B'D'$ is cyclic. This means that $AC'B'D'$ is an isosceles trapezoid. Now $K_1'$ is a circle tangent to $AD'$ and to $K'$. Furthermore, $K_1'$ is internally tangent to $(AC'B'D')$ at point $M'$. We know that $M'$ lies on arc $\overarc{B'D'}$ of $(AC'B'D')$. Note that $AC = MC$ is equivalent to $AM' = M'C'$, which in turn is equivalent to the statement that $M'$ lies on the perpendicular bisector of $AC'$. Furthermore, note that $AM$ bisects $\angle DAB$ is equivalent to $AM'$ bisects $\angle D'AB'$, which in turn is equivalent to the statement that $M'$ lies on the perpendicular bisector of $B'D'$. Let $P,Q$ denote the centers of $(AC'B'D'), K_1'$ respectively. Note that $M'$ is the intersection of ray $\overrightarrow{PQ}$ with $(AC'B'D')$. We want to show that $M'$ lies on the perpendicular bisector of $AC'$ if and only if it lies on the perpendicular bisector of $B'D'$. This is equivalent to showing that $Q$ lies on the perpendicular bisector of $AC'$ if and only if it lies on the perpendicular bisector of $B'D'$. Let $\ell$ denote the line parallel to $\overline{AD'}$ that is equidistant from $\overline{AD'}$ and $\overline{B'C'}$. Note that $\ell$ passes through $Q$ and the midpoints of $AC',B'D'$. Now if $Q$ lies on the perpendicular bisector of $AC'$, then $\ell$ must be perpendicular to $AC'$, implying that $AC'B'D'$ is a rectangle. From here, it follows that $\ell$ is also the perpendicular bisector of $B'D'$, so $Q$ is on the perpendicular bisector of $B'D'$. Similarly, if $Q$ lies on the perpendicular bisector of $B'D'$, then $\ell$ must be perpendicular to $B'D'$, implying that $AC'B'D'$ is a rectangle. From here, it follows that $\ell$ is also the perpendicular bisector of $AC'$, so $Q$ is on the perpendicular bisector of $AC'$. We have shown that the two given statements are equivalent, as desired. $\blacksquare$