Since $BFC$ is a right-angled triangle, we have $\angle BCF = 90^\circ - \beta$. As $CDH$ is also a right-angled triangle, we get $\angle DHC = \beta$.
It follows that $\angle AHF = \beta$ (because the angles $\angle DHC$ and $\angle AHF$ are vertically opposite).
As $ABKC$ is a cyclic quadrilateral, inscribed angles which subtend $\overline{AC}$ are equal to $\angle AKC = \angle ABC = \beta$. We have that $\angle ACK$ is a right angle because it subtends
the diameter $\overline{AK}$ (using Thales' theorem). We now infer that the triangles $AFH$ and $ACK$ are similar (because we know that $\angle AFH$ and $\angle ACK$ are right angles, whereas $\angle AKC = \angle AHF = \beta$). From this similarity, we get
$$\frac{|AF|}{|AH|} = \frac{|AC|}{|AK|}. \qquad (4)$$
We also know that $AFHE$ is a cyclic quadrilateral, because $\angle AFH + \angle HEA = 90^\circ + 90^\circ = 180^\circ$.
Therefore, $\angle FEA = \angle FHA = \beta$ holds, because both these angles subtend the chord $\overline{AF}$. Thus, triangle $AFE$ shares an angle ($\angle BAC = \angle FAE$) with triangle $ABC$. Since we also have $\angle FEA = \beta$, it follows that $AEF$ and $ABC$ are similar, so that
$$\frac{|AE|}{|AF|} = \frac{|AB|}{|AC|}. \qquad (5)$$
The quadrilateral $ABKC$ being cyclic, we know that the inscribed angles which subtend $\overline{AB}$ are equal to $\angle BKA = \angle BCA = \gamma$.
In triangle $AFE$ we have $\angle AFE = \gamma$ (by similarity with triangle $ABC$), so its complementary angle equals $\angle HFE = 90^\circ - \gamma$. Notice that $\angle HAE = \angle HFE = 90^\circ - \gamma$, because these are both inscribed angles inside the cyclic quadrilateral $AFHE$. It follows that the triangles $AGE$ and $AMB$ are similar, so that
$$\frac{|AG|}{|AE|} = \frac{|AM|}{|AB|}. \qquad (6)$$
By multiplying (4), (5) and (6) we obtain the equality of ratios $\frac{|AG|}{|AH|}$ and $\frac{|AM|}{|AK|}$.
Finally, using the converse to the intercept theorem, we conclude that the lines $GM$ and $HK$ are parallel.