A problem about orthocenter and circumscribed circle The orthocenter of $\triangle ABC$ is $H$. $E,F$ are on $BC, AC$ such that $\angle EHF = \angle C$. $G$ is on the circumscribed circle of $\triangle ABC$ such that $AG \parallel HE$. Prove that $EF$ bisects $HG$
I found the condition $\angle EHF = \angle C$ rather awkward but important. Not sure how to use it.. 

 A: It is clear that under the given circumstances $E$ determines $G$ (and $F$), and conversely $G$ determines $E$ (and $F$). So we can restate equivalently:

Let $\Delta ABC$ be a given triangle with orthocenter $H$, circumcenter $(O)$, and circumcircle $(O)=(ABC)$. Let $G$ be a point on $(O)$ (on the arc $\overset\frown{BC}$ not containing $A$). We draw the lines $GA$, $GB$, and parallels through $H$ to these lines, so that the pairs of parallels determine the parallelogram $GPHQ$, so that $HP\|AQG$, and $HQ\|BPG$. Let $E=HP\cap BC$. Let $F=HQ\cap AC$.

Let $X$ be the intersection of the diagonals in the parallelogram $GPHQ$.
Then the points $E,X,F$ are collinear.

Proof:
We will denote by $x$ the measure of the angles in $A,B$ that cover the arc $\overset \frown{GC}$. The orange angles are (in measure) equal to the angle in $C$ of the given triangle. We will write $\bar C$ for (the measure) of the complement angle, so $C+\bar C=\frac\pi2$.
Let us show first the equality of proportions:
$$
\tag{$*$}
\frac{EH}{EP}=
\frac{FH}{FQ}
$$
by using proportions of areas, and the formula for the area of a triangle, expressed as half of the product of two sides times the sine of the angle between them. We compute separately:
$$
\begin{aligned}
\frac{EH}{EP}
&=
\frac{[BEH]}{[BEP]}
=
\frac
{BE\cdot BH\;\sin \bar C}
{BE\cdot BP\;\sin x}
=
\frac{BH}{BP}\cdot\frac{\sin \bar C}{\sin x}
\\
&
=
\frac
{\sin \widehat{BPH}}
{\sin \widehat{BHP}}
\cdot\frac{\sin \bar C}{\sin x}
=
\frac
{\sin \widehat{BGA}}
{\sin \angle(BH,AG)}
\cdot\frac{\sin \bar C}{\sin x}
=
\frac{\sin C\cdot\sin \bar C}{\sin \bar x\cdot \sin x}
\ ,
\\[3mm]
\frac{FH}{FQ}
&=
\frac{[AFH]}{[AFQ]}
=
\frac
{AF\cdot AH\;\sin \bar C}
{AF\cdot AQ\;\sin x}
=
\frac{AH}{AQ}\cdot\frac{\sin \bar C}{\sin x}
\\
&
=
\frac
{\sin \widehat{AQH}}
{\sin \widehat{AHQ}}
\cdot\frac{\sin \bar C}{\sin x}
=
\frac
{\sin \widehat{AGB}}
{\sin \angle(AH,BG)}
\cdot\frac{\sin \bar C}{\sin x}
=
\frac{\sin C\cdot\sin \bar C}{\sin \bar x\cdot \sin x}\ .
\end{aligned}
$$
This shows $(*)$. Now we can apply the reciprocal of Menelaos' Theorem, applied for the triangle $\Delta HPQ$, with points $E\in HP$, $X\in PQ$, $F\in QH$ on its sides. The product
$$
\frac{EH}{EP}\cdot
\underbrace{\frac{XP}{XQ}}_{=1}\cdot
\frac{FQ}{FH}=1
\ ,
$$
using $(*)$,
so the points $E,X,F$ are collinear.
$\square$

(Strictly speaking, in the picture we should also consider signs, so $XP:XQ=-1$, and $EH:EP<0$, please arrange for correct signs, if this pedant point matters...)

Bonus: The homothety centered in $H$ with factor $1/2$ maps the circumcircle $(ABC)$ in the Euler circle $(A_1B_1C_1)$, where $A_1,B_1,C_1$ are the mid points of the segments $HA$, $HB$, $HC$. So
$G\in (ABC)$, is mapped into $X$, (the mid point of $HG$,) which is a point on the Euler circle.
