Problem 968

13. (1) As shown in Figure 7, connect $B O$ and extend it to intersect $\odot O$ at point $F$.

Since $O$ is the circumcenter of $\triangle A B C$, $B F$ is the diameter of $\odot O$.
Thus, $A F \perp A B$ and $F C \perp B C$.
Given that $H$ is the orthocenter of $\triangle A B C$, we have $H C \perp A B$.
Therefore, $A F \parallel H C$.
Similarly, $F C \parallel A H$.
Hence, quadrilateral $A H C F$ is a parallelogram,
$$F C = A H \text{. }$$

Draw $O M \perp B C$ at point $M$, then
$$O M = \frac{1}{2} F C \text{. }$$

Since $G$ is the midpoint of $A H$ and $G K = H D$, we have
$$\begin{array}{l} K D = K H + H D = K H + G K \\ = G H = \frac{1}{2} A H = \frac{1}{2} F C = O M \text{. } \end{array}$$

Given that $K D \parallel O M$, quadrilateral $O M D K$ is a rectangle.
Therefore, $O K \parallel M D$, which means $E K \parallel B C$.
(2) Draw $G N \perp A B$ at point $N$.

Since $H$ is the orthocenter of $\triangle A B C$,
$$\angle N A G = 90^{\circ} - \angle A B C = \angle D C H \text{. }$$

Given that $H D \perp B C$,
$\triangle A N G \sim \triangle C D H$
$$\Rightarrow \frac{N G}{D H} = \frac{A G}{C H}, \angle N G A = \angle D H C \text{. }$$

Since $G K = H D$ and $A G = G H$, we have
$$\begin{array}{c} \frac{N G}{G K} = \frac{G H}{H C} \text{. } \\ \text{Also, } \angle N G K = 180^{\circ} - \angle N G A \\ = 180^{\circ} - \angle D H C = \angle G H C \text{, } \end{array}$$

Thus, $\triangle N G K \sim \triangle G H C$
$$\Rightarrow \angle K N G = \angle C G H \text{. }$$

From (1), we know $G K \perp K E$.
Therefore, points $E$, $K$, $G$, and $N$ are concyclic
$$\begin{array}{l} \Rightarrow \angle C G H = \angle K N G = \angle G E K \\ \Rightarrow \angle E G C = \angle E G K + \angle C G H \\ \quad = \angle E G K + \angle G E K = 90^{\circ} \\ \Rightarrow G E \perp G C \text{. } \end{array}$$