First Solution: (Based on work by Ryan Ko) Let $M$ be the midpoint of segment $AH$. Since $\angle AEH = \angle AFH = 90^\circ$, quadrilateral $AEHF$ is cyclic with $M$ as its circumcenter. Hence triangle $EFM$ is isosceles with vertex angle $\angle EMF = 2\angle CAB = 2x$. Likewise, triangle $PQO$ is also an isosceles angle with vertex angle $\angle POQ = 2x$. Therefore, triangles $EFM$ and $PQO$ are similar.

Since $AEHF$ and $APHQ$ are cyclic, we have $\angle EFH = \angle EAH = \angle PQH$ and $\angle FEH = \angle FAH = \angle QPH$. Consequently, triangles $HEF$ and $HPQ$ are similar. It is not difficult to see that quadrilaterals $EHFM$ and $PHQO$ are similar. More precisely, if $\angle QHF = \theta$, there is a spiral similarity $S$, centered at $H$ with clockwise rotation angle $\theta$ and ratio $QH/FH$, that sends $FMEH$ to $QOPH$. Let $R_1$ be the point in between $B$ and $D$ such that $\angle R_1HD = \theta$. Then triangles $QHF$ and $R_1HD$ are similar. Hence $S(D) = R_1$. It follows that
$$S(DFME) = R_1QOP.$$
It is well known that points $D$, $E$, $F$, and $M$ lie on a circle (the nine-point circle of triangle $ABC$). (This fact can be established easily by noting that $ABDE$ and $ACDF$ are cyclic, implying that
$\angle FDB = \angle CAF = x$, $\angle EDC = \angle BAE = x$, and $\angle EDF = 180^\circ - 2x = 180^\circ - \angle EMF$.) Since $DFME$ is cyclic, $R_1QOP$ must also be cyclic. By the given conditions of the problem, we conclude that $R_1 = R$, implying that
$$S(DEF) = RPQ,$$
or triangles $DEF$ and $RPQ$ are similar. It follows that
$$\frac{ED}{FD} = \frac{PR}{QR}.$$

Now we are ready to finish our proof. Since $ACDF$ and $ABDE$ are cyclic, $\angle BFD = \angle AFE = \angle ACB = z$. Thus $\angle DFE = 180^\circ - 2z$. Since triangles $DEF$ and $RPQ$ are similar, $\angle RQP = 180^\circ - 2z$. Because $CR$ is tangent to the circumcircle of triangle $PQR$, $\angle CRP = \angle RQP = 180^\circ - 2z$. Thus, in triangle $CPR$, $\angle CPR = z$, and so it is isosceles with $CR = PR$. Likewise, we have $BR = QR$. Therefore, we have
$$\frac{ED}{FD} = \frac{PR}{QR} = \frac{CR}{BR}.$$

Second Solution: (Based on work by Zarathustra Brady) Let the circumcircle of triangle $BQH$ meet line $BC$ at $R_3$ (other than $B$).

Since $APHQ$ and $BQHR_3$ are cyclic, $\angle PHQ = 180^\circ - \angle PAQ$ and $\angle QHR_3 = 180^\circ - \angle QBR_3$, implying that $\angle PHR_3 = 360^\circ - \angle PHQ - \angle QHR_3 = 180^\circ - \angle ACB$. Hence $CPHR_3$ is also cyclic.

(We just established a special case of Miquel's Theorem.) Because $BQHR_3$ and $CR_3HP$ are cyclic, we have $\angle QR_3H = \angle QBH = 90^\circ - \angle BAC$ and $\angle HR_3P = \angle HCP = 90^\circ - \angle BAC$. Hence $\angle QR_3P = 180^\circ - 2\angle BAC = 180^\circ - 2x$. Likewise, we have $\angle PQR = 180^\circ - 2z$ and $\angle R_3PQ = 180^\circ - 2y$. As we have shown in the first solution, triangle $DEF$ have the same angles. Hence triangle $R_3PQ$ is similar to triangle $DEF$. Also note that $\angle POQ + \angle PR_3Q = 2x + 180^\circ - 2x = 180^\circ$, implying that $R_3$ lies on the circumcircle of triangle $OPQ$. By the given condition, have $R_3 = R$. We can then finish our proof as we did in the first solution.

$$\frac{ED}{FD} = \frac{PR}{QR} = \frac{CR}{BR}.$$