IRL ABooklet

Because $AN$ is the perpendicular bisector of $BM$, triangle $ABM$ is isosceles with $\angle ABM = \angle AMB$. Similarly, $\angle ECB = \angle EBC$ in the isosceles triangle $EBC$. We have $\angle AMB = \angle MAC + \angle ECB$ (external angle) and $\angle ABM = \angle ABE + \angle EBC$.

Combining these equalities, we obtain
$$\angle ABE + \angle EBC = \angle MAC + \angle ECB = \angle MAC + \angle EBC,$$
hence $\angle ABE = \angle MAC$ which means that $AC$ is tangent to the circumcircle of $\triangle ADB$. This proves (a).

To prove (b) we note that $\triangle ACB \sim \triangle DBM$ since $\angle ABM = \angle AMB$ and $\angle ECB = \angle EBC$. Consequently,
$$\frac{|DM|}{|AM|} = \frac{|DM|}{|AB|} = \frac{|BM|}{|CB|} = \frac{1}{2}$$
which means that $D$ is the midpoint of $AM$.

Let $G$ be the intersection of $AN$ and $BE$. Since $GN \parallel EM$, $|EM|/|GN| = |BM|/|BN| = 2$. On the other hand, as $EM$ and $AN$ are parallel, we have
$$\frac{|AN|}{|EM|} = \frac{|CN|}{|CM|} = \frac{3}{2},$$
hence $|AN| = \frac{3}{2}|EM| = 3|GN|$. This shows that $G$ is the centroid of the isosceles triangle $ABM$ and $BG$ intersects $AM$ at its midpoint. This proves part (b).

From our calculation above we get $|AG| = \frac{2}{3}|AN| = |EM|$ thus $AGME$ is a parallelogram. This implies that $\angle EAD = \angle AMG$. From the symmetry of the isosceles triangle $ABM$ we obtain $\angle AMG = \angle ABD$, hence $\angle EAD = \angle ABD$ and $AC$ is tangent to the circumcircle of triangle $ADB$.