Problem 1021

Proof: As shown in the figure, let $\angle ABC = \beta, \angle BCA = \gamma$.

From $\angle BFC = 90^{\circ}$, we know $\angle BCF = 90^{\circ} - R$.

Since $\angle HDC = 90^{\circ}$, then $\angle DHC = \beta$.

Because $\angle DHC$ and $\angle AHF$ are vertical angles, so $\angle AHF = \beta$.
From the fact that $A, B, K, C$ are concyclic, we have
$$\angle AKC = \angle ABC = \beta, \angle BKA = \angle BCA = \gamma.$$

Since $AK$ is the diameter, $\angle ACK = 90^{\circ}$.
From $\angle AFH = \angle ACK = 90^{\circ}$, and $\angle AKC = \angle AHF = \beta$,
then $\triangle AFH \sim \triangle ACK \Rightarrow \frac{AF}{AH} = \frac{AC}{AK}$
(20 points)
From $\angle AFH + \angle HEA = 180^{\circ} \Rightarrow A, F, H, E$ are concyclic $\Rightarrow \angle FEA = \angle FHA = \beta$.
From $\angle BAC = \angle FAC \Rightarrow \triangle AEF \sim \triangle ABC \Rightarrow \frac{AE}{AF} = \frac{AB}{AC}$
In $\triangle AFE$, $\angle AFE = \gamma$, then $\angle HFE = 90^{\circ} - \gamma$.
Notice that, $\angle HAE = \angle HFE = 90^{\circ}$
Thus $\triangle AGE \sim \triangle AMB \Rightarrow \frac{AG}{AE} = \frac{AM}{AB}$.
(40 points)
From $\frac{AF}{AH} = \frac{AC}{AK}, \frac{AE}{AF} = \frac{AB}{AC}, \frac{AG}{AE} = \frac{AM}{AB}$
we get $\frac{AG}{AH} = \frac{AM}{AK}$
Therefore $GM \parallel HK$.
(50 points)