Angles in nine-point circle scenario

Question

Let the sides $AD$ and $BC$ of the quadrilateral $ABCD$ (such that $AB$ is not parallel to $CD$) intersect at point $P$. Points $O_1$ and $O_2$ are the circumcenters and points $H_1$ and $H_2$ are the orthocenters of triangles $ABP$ and $DCP$, respectively. Denote the midpoints of segments $O_1 H_1$ and $O_2 H_2$ by $E_1$ and $E_2$, respectively. Prove that the perpendicular from $E_1$ on $CD$, the perpendicular from $E_2$ on $AB$ and the line $H_1 H_2$ are concurrent.

Solution 2. Let the perpendicular from $E_1$ on $CD$ meet $PH_1$ at $X$, and the perpendicular from $E_2$ on $AB$ meet $PH_2$ at $Y$ (see Figure 3). Let $φ$ be the intersection angle of $AB$ and $CD$. Denote by $M$, $N$ the midpoints of $PH_1$m $PH_2$, respectively.

We will prove now that triangles $EX_1M$ and $E_2YN$ have equal angles at $E_1$, $E_2$, and supplementary angles at $X$, $Y$.

In the following, angles are understood as oriented, and equalities of angles modulo $180°$.

Let $α = ∠H_2PD$, $ψ = ∠DPC$, $β = ∠CPH_1$. Then $α + ψ + β = φ$, $∠E_1XH_1 = ∠H_2YE_2 = φ$, thus $∠MXE_1 + ∠NYE_1 = 180°$.

By considering the Feuerbach circle of $\mathord{△}ABP$ whose center is $E_1$ and which goes through $M$, we have $∠E_1MH_1 = ψ + 2β$. Analogous consideration with the Feuerbach circle of $\mathord{△}DCP$ yield $∠H_2NE_2 = ψ + 2α$. Hence indeed $∠XE_1M = φ - (ψ + 2β) = (ψ + 2α) - φ = ∠YE_2N$.

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This is from IMO'09 (full pdf is here: http://www.imo-official.org/problems/IMO2009SL.pdf). Could someone please explain why angles $E_1MH_1$ and $H_2NE_2$ are what they are in the last paragraph?

Answer 1

$\def\tri{\mathord{△}}$It suffices to consider $\tri PAB$. In the figure above, $K$ is the perpendicular foot of $B$ on line $AP$, and $S$ is the midpoint of segment $BH$. Note that circle $KSM$ is the nine-point circle of $\tri PAB$. Since $MS$ is parallel to $PB$, then$$ ∠SMH = ∠BPH = β. $$ Because $E$ is the center of circle $KSM$. then$$ ∠EMS = \frac{π}{2} - \frac{1}{2} ∠SEM = \frac{π}{2} - ∠SKM = \frac{π}{2} - ∠KHM = ∠KPH = ψ + β. $$ Thus,$$ ∠EMH = ∠EMS + ∠SMH = ψ + 2β. $$