The cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ shown in Figure 2 is cut into two parts by a plane passing through point $A$, the point $H$ which is the closer one-third point of edge $C C^{\prime}$ to $C$, and the midpoint $F$ of edge $D D^{\prime}$. Determine the ratio of the volumes of the two parts.
Let the plane that bisects the cube be denoted by $S$, and the intersection point of line $BC$ and plane $S$ be denoted by $G$. We choose the edge length of the cube to be 6 units. The lines $AF$ and $GH$ lie in the plane $S$, but they cannot intersect each other because they are in the planes $ADD'A'$ and $BCC'B'$, which are two parallel face planes of the cube. Therefore, $AF$ and $GH$ are parallel. Thus, the triangles $AFD$ and $GHC$ are similar, from which we get that

$$
GC = CH \cdot \frac{AD}{DF} = 2 \cdot \frac{6}{3} = 4
$$

Let $P$ be the point on the ray $DC$ that is 18 units away from $D$. Then

$$
\frac{FD}{HC} = \frac{DP}{CP} = \frac{AD}{GC}
$$

and since $FD \parallel HC$ and $AD \parallel GC$, it follows from the intercept theorem that the lines $FH$ and $AG$ also pass through $P$. This means that the solid $AFDGHC$ is a frustum, with bases $AFD$ and $GHC$, and height $DC$; therefore, its volume, according to the known formula, is

$$
V_{1} = \frac{DC}{3} \left( T_{AFD} + \sqrt{T_{AFD} \cdot T_{GHC}} + T_{GHC} \right) = 2 \cdot (9 + 6 + 4) = 38
$$

Since the volume of the other part of the cube is $V_{2} = 6^3 - V_{1} = 178$, the ratio of the volumes of the two parts separated by the plane $S$ is $38:178 = 19:89$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f0f80c20df9d2b233149g-1.jpg?height=336&width=718&top_left_y=1060&top_left_x=693)