Consider Figure 2:

Figure 2

We note that $H$ must also be on $V$, the circumcircle of triangle $BDC$. This is because triangles $BLH$ and $BMD$ are similar (note that triangle $ABD$ is iscoceles, with $BA = BD$, and $BM$ is a perpendicular bisector of $AD$, that also bisects $\angle ABD$), implying that angles $BHC$ and $BDC$ are equal, showing that $BHDC$ is a cyclic quadrilateral.

Let $O$ be the centre of $W$, the circumcircle of triangle $HDF$ and drop the perpendicular from $O$ to $DF$, with foot $P$. Put $\theta = \angle DOP$, so that $\angle DOF = 2\theta$, giving $\angle DHF = \theta$. Since $HM$ is a perpendicular bisector of $AD$, and $HA = HD$, it follows that $\angle HAC = 90^\circ - \theta$. Thus, $\angle ACN = \theta$. But the chords $BD$ and $BE$ of circle $V$ have equal length (since they are both radii of circle $K$), hence subtend the same angle in circle $V$, giving $\angle BCE = \theta$. But then $\angle BDE = \theta$, so that $\angle FDR = \theta$, where $R$ is an arbitrary point on the line through $B$ and $D$, with $R$ and $B$ on opposites sides of $D$.

Finally, since $\angle ODP = 90^\circ - \theta$, we conclude that $\angle ODR = 90^\circ$, i.e. the line through $B$ and $D$ must be tangent to $W$, at the point $D$.