VMO

a) Firstly, note that when $\ell$ changes, the angles $\angle AMN$ and $\angle ANM$ both remain unchanged, so triangle $AMN$ is always self-congruent. Draw $AK$ perpendicular to $MN$ ($K\in MN$), then $AK\le AH$. Therefore, the area of triangle $AMN$ attains maximal value when $AH$ is the altitude or $MN\perp AH$. Thus, when $\Delta \perp AH$, the area of triangle $AMN$ is largest.



b) Let the line passing through $A$ and parallel to $BD$ intersect (ABM) at $E$ and the line passing through $A$ and parallel to $CD$ intersect (ACH) at $F$. Notice that the radius of the circles ($O$), (ABH) and (ACH) are equal, hence $\angle ADB=\angle AEB$, which implies $\angle ABD=\angle BAE$ (because $BD\parallel AE$). Hence, $\angle BAD=\angle ABE$, it follows that $AD\parallel BE$. Thus $ADBE$ is a parallelogram.

Similarly, $ADCF$ is a parallelogram. Therefore, triangle $AEF$ is the image of triangle $DBC$ through the translation $T_v$ with vector $\vec{v}=\overrightarrow{DA}$. Denote $O_1,H_1$ to be the circumcenter and orthocenter of triangle $AEF$, respectively. It is well known that $$\begin{array}{rl}\overrightarrow{OH}& =\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\\ \overrightarrow{O_1{H}_1}& =\overrightarrow{O_{1}A}+\overrightarrow{O_{1}E}+\overrightarrow{O_{1}F},\end{array}$$ However, $\overrightarrow{O_1{H}_1}=\overrightarrow{O_{1}O}+\overrightarrow{O{H}_1}$, $\overrightarrow{O_{1}A}=\overrightarrow{O_{1}O}+\overrightarrow{OA}$, and $\overrightarrow{O_{1}E}=\overrightarrow{OB}$, $\overrightarrow{O_{1}F}=\overrightarrow{OC}$ (image via translation forward $T_v$). Hence $$\overrightarrow{O_{1}O}+\overrightarrow{O{H}_1}=\overrightarrow{O_{1}O}+\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC},$$ so $\overrightarrow{O{H}_1}=\overrightarrow{OH}$. Thus, $H_1\equiv H$.



Now we will prove that the intersection $P$ of $d_1$ and $d_2$ always lies on the circle ($AEF$). First, since $AE\parallel BD$ and $AF\parallel CD$, we have $MP\perp AE$ and $NP\perp AF$. Let $P_1$ be the point of symmetry of $M$ through $AE$, $P_2$ is the symmetry point of $N$ over $AF$. Since the circles ($AEF$), ($ABH$) and ($ACH$) are equal (because they are equal to the same circle ($O$)) so $P_1,P_2$ are all on the circle ($AEF$). Note that the Steiner lines of $P_1$ and $P_2$ with respect to the triangle $AEF$ are coincident (that is the line $\Delta$). According to the property of the Steiner line, we deduce that $P_1\equiv P_2\equiv P$ so $P$ always lies on the fixed circle ($AEF$). $\square$