Consider a triangle $ABC$ with orthocenter $H$ and altitudes $AD, BE, CF$. Let the line FE meet the circumcircle of $ABC$ at a point P which does not lie on the same arc $AC$ as $B$. Let the intersection between $FD$ and $BP$ be $Q$.
Prove that $HQEP$, $QDCP$, $AHQB$ are cyclic.
Diagram:
The line $EF$ intersects the circumcircle $k$ of triangle $ABC$ in two points. One of them is $P$, let the other be $G$. Then $ \angle \, BFG = \angle \, AFE = \angle \, BCA = \gamma = \angle \, BFD$. Observe that $$\gamma = \angle \, BCA = \frac{1}{2} \, arc(BGA) = \frac{1}{2} \, ( arc(AG) + arc(BG))$$ and that $$\gamma = \angle \, BFG = \frac{1}{2} \, ( arc(AP) + arc(BG)) $$ which implies that $arc(AG) = arc(AP)$ and thus $$AG = AP$$ It also means that $BA$ is the angle bisector of angle $\angle \, GBP$ and thus $$\angle \, FBG = \angle \, ABG = \angle \, ABP = \angle \, ABQ$$ Now, we can conclude that triangles $FBG$ and $FBQ$ are congruent because $BF$ is a common segment, $\angle \, FBG = \angle \, FBQ$ and $\angle \, BFG = \angle \, BFQ = \gamma$. Consequently, $BG = BQ$ and so triangles $ABG$ and $AQB$ are congruent because $AB$ is a common edge, $\angle \, ABG = \angle \, ABQ$ and $BG = BQ$. Therefore $$AG = AQ = AP$$ and so if we construct the circle $k_A$ centered at points $A$ and of radius $AG$, it passes through the points $G, \, P$ and $Q$.
Now, the inversion with respect to circle $k_A$ maps line $GP$ to circle $k$ and so points $E$ and $F$ are mapped to points $C$ and $B$ respectively. Furthermore, the circles circumscribed around the quads $BFHD$ and $CEHD$ are mapped by the inversion to themselves (i.e. they are orthogonal to $k_A$) because $F$ is mapped to $B$ and $E$ is mapped to $C$ and thus point $H$ is mapped to point $D$. For that reason line $DF$ is mapped to the circle through points $A, \, H$ and $B$ and since $FD \cap k_A = Q$ and $Q$ is mapped to itself, the quad $AHQB$ is inscribed in a circle.
Finally, it is easy to see that quad $QDCP$ is inscribed in a circle. Indeed, $$\angle\, CPQ = \angle \,CPB = \angle \, CAB = \alpha = \angle \, BDF = \angle \, BDQ$$ Hence, the inverse image of $QDCP$ with respect to $k_A$ is $QHEP$ so $QHEP$ should also be inscribed in a circle.
By “angles in the same segment CPAB”, we have $\angle Red_1 = \angle Red_2$. By “exterior angle of cyclic quadrilateral FACD”, we have $\angle Red_1 = \angle Red_3$. This means QDCP is cyclic.
Similarly, by considering the purple marked angles, we get AFQP being cyclic also.
Note that $DEF$ is the corresponding orthic triangle of $\triangle ABC$. The orthocenter (H) of ⊿ABC is also the in-center of ⊿DEF. Therefore, we have HF bisects $\angle PFD$. This, together with AD being the altitude, makes $\angle Purple_1 = \angle Purple_2$.
In addition, from the fact that AFQP is cyclic, we have $\angle Purple_1= \angle Purple_3$ and $\angle Purple_2 = \angle Purple_4$. Therefore $\triangle APQ$ is isosceles and its base angles are purple marked.
Construction: BHE is extended to cut the circle ACB at Y. A and Y are then joined.
Note that $\angle Purple_6 = \angle Purple_0 = \angle Purple_5$. This makes $\triangle AYH$ is also isosceles with base angles are purple marked.
After transferring the above info to the figure below, I think it should be clear that $\alpha = \beta$. Since $\beta = \delta$ also, we have ABQH being cyclic.
The rest is easy.