Kanada

Let $R$ denote the intersection of lines $ED$ and $PH$. Since quadrilaterals $ECDH$ and $APHB$ are cyclic, we have $\angle RDA = 180^\circ - \angle EDA = 180^\circ - \angle EDH = 180^\circ - \angle ECH = 90^\circ + A$, and $\angle RPA = \angle HPA = 180^\circ - \angle HBA = 90^\circ + A$. Therefore, $APDR$ is cyclic. This in turn implies that $\angle PBE = \angle PBH = \angle PAH = \angle PAD = \angle PRD = \angle PRE$, and so $PBRE$ is also cyclic.
Let $F$ denote the base of the altitude from $C$ to $AB$. Then $D, E, F$, and $M$ all lie on the 9-point circle of $\triangle ABC$, and so are cyclic. We also know $APDR, PBRE, BCEF$, and $ACDF$ are cyclic, which implies $\angle ARB = \angle PRB - \angle PRA = \angle PEB - \angle PDA = \angle PEF + \angle FEB - \angle PDF + \angle ADF = \angle FEB + \angle ADF = \angle FCB + \angle ACF = C$. Therefore, $R$ lies on the circumcircle of $\triangle ABC$.
Now let $Q'$ and $R'$ denote the intersections of line $MQ$ with the circumcircle of $\triangle ABC$, chosen so that $Q', M, Q, R'$ lie on the line in that order. We will show that $R' = R$, which will complete the proof. However, first note that the circumcircle of $\triangle ABC$ has radius $\frac{AB}{2\sin C}$, and the circumcircle of $\triangle ABH$ has radius $\frac{AB}{2\sin\angle AHB} = \frac{AB}{2\sin(180^\circ - C)}$. Thus the two circles have equal radius, and so they must be symmetrical about the point $M$. In particular, $MQ = MQ'$.
Since $\angle AEB = \angle ADB = 90^\circ$, we furthermore know that $M$ is the circumcenter of both $\triangle AEB$ and $\triangle ADB$. Thus, $MA = ME = MD = MB$. By Power of a Point, we then have $MQ \cdot MR' = MQ' \cdot MR' = MA \cdot MB = MD^2$. In particular, this means that the circumcircle of

$\triangle DR'Q$ is tangent to $MD$ at $D$, which means $\angle MR'D = \angle MDQ$.
Similarly $MQ \cdot MR' = ME^2$, and so $\angle MR'E = \angle MEQ = \angle MDQ = \angle MR'D$. Therefore, $R'$ also lies on the line $ED$.
Finally, the same argument shows that $MP$ also intersects the circum-
circle of $\triangle ABC$ at a point $R''$ on line $ED$. Thus, $R, R',$ and $R''$ are
all chosen from the intersection of the circumcircle of $\triangle ABC$ and the
line $ED$. In particular, two of $R, R',$ and $R''$ must be equal. However,
$R'' \neq R$ since $MP$ and $PH$ already intersect at $P$, and $R'' \neq R'$ since
$MP$ and $MQ$ already intersect at $M$. Thus, $R' = R$, and the proof is
complete. $\square$