BMO, 2017

WLOG, assume $AB < AC$. Let $M$ be the midpoint of side $BC$ and let the circumcircle of $DFE$ intersect $AF$ again at $K$. Since
$$90^\circ + \angle MED = 180^\circ - \angle MDE = \angle ABC + \frac{\angle BAC}{2} = \angle AFE = \angle DFE + \angle AFD = 90^\circ + \angle AFD$$
it follows that
$$\angle AFD = \frac{\angle ABC - \angle ACB}{2} = \angle MED$$
Because
$$\angle DKE = \angle DME = 90^\circ$$
and
$$\angle KED = \angle KFE = \angle MED$$
we get $\triangle KDE \cong \triangle MDE$ from which it follows that $DE$ is the perpendicular bisector of $MK$ and here we get
$$\angle FAD = \angle KAD = \angle MAD$$
It is obvious that $P$ is the intersection of $ME$ and $AF$. Let $ME$ intersect $\Gamma$ again at $L$. From the angle bisector theorem in triangle $\triangle AMP$ we get
$$\frac{PE}{ME} = \frac{AP}{AM} = \frac{LP}{LM} \quad (1)$$
($LA$ is the external angle bisector of $\angle MAP$ since $LE$ is the diameter of $\Gamma$). Now we prove that $CE$ and $CL$ are angle bisectors of $\angle MCP$. Let $M'$ be the point on $LE$ such that $\angle M'CE = \angle ECP$. From the angle bisector theorem we get
$$\frac{M'E}{PE} = \frac{CM'}{CP} = \frac{M'L}{PL} \quad (2)$$
Multiplying (1) and (2) we get $\frac{ME}{LM} = \frac{M'E}{M'L}$ adding 1 on both sides we get $LM = LM'$ from which it follows that $M = M'$ and thus $CE$ and $CL$ are the bisectors of $\angle MCP$. Now we have
$$\angle MPC = 90^\circ - \angle MCP = 90^\circ - 2\angle MCE = 90^\circ - 2\angle EAC = 90^\circ - \angle BAC$$

Since $X$ and $Y$ are perpendicular to $AB$ and $AC$ we have $BXPM$ and $CYPM$ are concyclic. Here we get
$$\angle MYC = \angle MPC = 90^\circ - \angle BAC$$
and it follows that $YM \perp AX$. Similarly we get $XM \perp AY$ and so $M$ is the ortocenter of $\triangle AXY$ giving us $M = H'$.
Since $ATHS$ and $ATQF$ are both concyclic it is enough to prove that $HSFQ$ is concyclic. Since
$$\angle BQC = 180^\circ - \angle BAC = \angle BHC$$
and $HQ \perp BC$ it follows that $BC$ is the perpendicular bisector of $HQ$. It is enough to prove that $BC$ is the perpendicular bisector of $SF$. Let $AM$ and $TH$ meet $\Gamma$ again at points $A'$ and $N$ respectively.
Since $HN$ passes through the midpoint of side $BC$ and
$$\angle BHC = 180^\circ - \angle BAC = \angle BNC$$
it follows that $BNCH$ is a parallelogram. From here we get that
$$\angle NCB = \angle HBC = 90^\circ - \angle ACB$$
giving us $\angle NCA = 90^\circ$ and similarly $\angle NBA = 90^\circ$. This means $AN$ is the diameter of $\Gamma$, so
$$\angle NA'A' = \angle NA'A = 90^\circ = \angle HSA = \angle HSA'$$
and from here we have $HS \parallel A'N$. Now since $HS \parallel A'N$ and $M$ is the midpoint of $HN$ (because $BHCN$ is a parallelogram) we get

that $HSNA'$ is a paralelogram. Since
$$\angle FAE = \angle EAM = \angle EAA'$$

HS, AT are concurrent.
we get that $FA'BC$ is an isosceles trapezoid which means that $ME$ is the perpendicular bisector of $FA'$ (since it is the perpendicular bisector of $BC$).
This gives us $BF = CA' = BS$ and $CF = BA' = CS$ giving us that $SBFC$ is a deltoid, meaning that $BC$ is the perpendicular bisector of $FS$. This means that $HSFQ$ is an isosceles trapezoid. Now from the radical axis theorem of the circumcircles of $HSFQ$, $HSAT$ and $ATQF$ we get that $QF$,