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 $△ADB$. This proves (a).

To prove (b) we note that $△ACB\sim △DBM$ 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 $AM$.

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Alternative solution.

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 $ADB$.