Solution:

We show first that $\mathrm{O}$ is the circumcenter of $\mathrm{ABC}$. $\angle \mathrm{ABF} = 90^\circ - \mathrm{A}$. The line $\mathrm{BC}$ is the reflection in $BD$ of the line $BA$ and the line $\mathrm{BF}'$ is the reflection of $\mathrm{BF}$, so angle $\mathrm{CBF}' = 90^\circ - \mathrm{A}$. But if $\mathrm{O}'$ is the circumcenter, then $\angle \mathrm{BO}'\mathrm{C} = 2 \angle \mathrm{BAC} = 2\mathrm{A}$, so $\angle \mathrm{O}'\mathrm{BC} = 90^\circ - \mathrm{A}$. Hence $\mathrm{O}'$ lies on $\mathrm{BF}'$. Similarly, it lies on $\mathrm{AE}'$ (the reflection of $\mathrm{AE}$ in the angle bisector of $\mathrm{A}$). Hence $\mathrm{O} = \mathrm{O}'$.

$\angle \mathrm{MBC} = \angle \mathrm{MAC} = 90^\circ - \mathrm{C}$ (since $\mathrm{AH}$ is an altitude) $= \angle \mathrm{FBC}$ (since $\mathrm{BF}$ is an altitude) $= \angle \mathrm{HBC}$ (same angle). So triangles HBE and MBE are congruent and $\mathrm{HE} = \mathrm{EM}$. [Note: this should be a familiar result.]
$\mathrm{AN}$ is a diameter, so angle $\mathrm{AMN} = 90^\circ = \angle \mathrm{AEC}$, so $\mathrm{BC}$ and $\mathrm{MN}$ are parallel. Hence $\mathrm{P}$ is the midpoint of $\mathrm{HN}$ and of $\mathrm{BC}$. So $\mathrm{OP}$ is perpendicular to $\mathrm{BC}$. So $\mathrm{AH}$ and $\mathrm{OS}$ are parallel.
Since $\mathrm{R}$ lies on $\mathrm{BC}$, triangles HER and MER are congruent, so $\angle \mathrm{EHR} = \angle \mathrm{EMR} = \angle \mathrm{AMO}$ (same angle) $= \angle \mathrm{MAO}$. Hence HS and AO are parallel. So AHSO is a parallelogram.