IRL ABooklet

(a) By construction, $M$, $N$ are on $\Gamma = (ABC)$ and $D$, $E$ are on $(ALM)$. The circles $\Gamma = (ABC)$ and $(BCD)$ have radical axis $BC$, and the circles $(ABC)$ and $(ALM)$ have radical axis $AM$.
Since $S$ is the intersection of these two radical axes, this point is the radical centre of the three circles $(ABC)$, $(BCD)$ and $(ALM)$. Thus $(BCD)$ and $(ALM)$ have radical axis $DS$. Since $DS$ intersects $(ALM)$ at $D$ and $E$, it follows that $E$ is also on $(BCD)$.

Let $Z$ be the intersection point of $NA$ and $BC$. Because $MN$ is a perpendicular bisector of a chord of $(ABC)$, it is a diameter of this circle, hence $\angle MAN = 90^\circ$ and so $\angle ZAM = 90^\circ$.
Since $ML$ is the perpendicular bisector of $BC$, $\angle ZLM = \angle BLM = 90^\circ$. Now $\angle ZAM = \angle ZLM = 90^\circ$ implies that $Z$ is on $(ALM)$ and $MZ$ is a diameter of this circle.
Moreover, since $MA$ is perpendicular to $NZ$ and $ZL$ is perpendicular to $MN$, as we have seen above, their intersection point $S$ is the orthocentre of $\triangle MNZ$, hence $NS$ is perpendicular to $ZM$. This means that $ZM$, being a diameter of $(ALM)$, is the perpendicular bisector of $DE$ which is a common chord of $(ALM)$ and $(BCD)$. Because $MN$ is the perpendicular bisector of $BC$ it follows that $M$ is the centre of $(BCD)$.

(b) Because $AM$ is perpendicular to $AN$ and $ZM$ is perpendicular to $NE$, we have $\angle AMZ = \angle AND$. In $(ALM)$ we have $\angle ADZ = \angle AMZ$ hence $ZD$ is tangent to $(ADN)$ (the alternate segment theorem). As $ZM$ is diameter of $(ALM)$, we have $\angle MDZ = 90^\circ$, so $ZD$ is also tangent to $BCD$ the centre of which was shown to be $M$ in part (a).

(a) Because $MN$ is the perpendicular bisector of $BC$, $MN$ is a diameter of $(ABC)$. Hence $\angle MCN = 90^\circ = \angle MLC$, and thus $\triangle MLC \sim \triangle MCN$, as they also share an angle at $M$. Hence
$$\frac{|ML|}{|MC|} = \frac{|MC|}{|MN|} \implies |MC|^2 = |ML| \cdot |MN|. \quad (25)$$
Also, the quadrilateral $ANLS$ is cyclic since $\angle NAM = \angle NLB = 90^\circ$. Since both $ANLS$ and $ADLM$ are cyclic, we have
$$\angle LNS = \angle LAM = \angle LDM$$
and thus $\triangle MLD \sim \triangle MDN$ (also sharing the angle at $M$), which gives
$$\frac{|ML|}{|MD|} = \frac{|MD|}{|MN|} \implies |MD|^2 = |ML| \cdot |MN|. \quad (26)$$
Equations (25) and (26) give $|MD| = |MC|$. As $M$ is on the perpendicular bisector of $BC$, we also have $|MC| = |MB|$, hence $M$ is the circumcentre of $\triangle BDC$.

(b) Let $Q$ be the circumcentre of $\triangle DNA$. It suffices to prove that $Q$, $D$, $M$ are collinear. On the one hand, for the circle ($ADN$):
$$\angle QDN = 90^\circ - \frac{1}{2} \angle DQN = 90^\circ - \angle NAD = \angle DAM. \quad (27)$$
On the other hand, the power of the point $M$ with respect to ($SLNA$) gives
$$|ML| \cdot |MN| = |MS| \cdot |MA|, \quad (28)$$
which together with equation (26) gives
$$|MD|^2 = |MS| \cdot |MA| \quad \text{and so} \quad \frac{|MS|}{|MD|} = \frac{|MD|}{|MA|}, \quad (29)$$
thus $\triangle MDA \sim \triangle MSD$, which implies
$$\angle DAM = \angle MDS. \quad (30)$$
Finally equations (27) and (30) give $\angle QDN = \angle MDS$, hence $M$, $D$, $Q$ are collinear since $N$, $D$, $S$ are.