Consider triangle $\triangle ABC$ with altitudes $AD$, $BF$, and $CE$. The orthocenter of the triangle is denoted by $H$. A line through $D$ that is parallel to $AB$ intersects line $EF$ at point $G$.

To find the angle $\angle CGH$, follow these steps:

1. **Identify the orthocenter $H$:**
Since $AD$, $BF$, and $CE$ are altitudes of $\triangle ABC$, they meet at the orthocenter $H$ of the triangle.

2. Analyze parallelism:
The line passing through $D$ and parallel to $AB$, when intersecting $EF$ at $G$, means that $DG \parallel AB$.

3. Use properties of cyclic quadrilaterals:
The points $A$, $B$, $C$, and $H$ are concyclic in the circumcircle. The key insight is noticing the properties of angles formed by such a configuration:
- Since $AB \parallel DG$, the angles $\angle DAG = \angle ABC$ are equal.
- Consider the quadrilateral $BFDG$: since it is cyclic, the opposite angles are supplementary.

4. **Calculate $\angle CGH$:**
- Since $H$ lies on the altitude $AD$, $\angle CDH = 90^\circ$.
- Now observe the angles formed at $G$:
- Since $DG \parallel AB$ and both are perpendicular to $CE$, we have $\angle CGH = 90^\circ$.

5. Conclusion:
This analysis ensures that $\angle CGH$ is a right angle since $G$ is where the line parallel to $AB$ meets $EF$ and forms perpendicularity with $CE$.

Thus, the angle $\angle CGH$ is:
$$\boxed{90^\circ}$$