4. Condition: A certain farmer has created a system of channels $A B C D E F G H$, as shown in the diagram. All channels are the same, and their connections occur at nodes marked by the letters $B, D$, and $G$. Water enters at node $A$ and exits at node $E$. Let's call the flow rate the number of cubic meters of water passing through a cross-section of the channel per unit of time.

The farmer discovered that when moving through the channels from one point $A, B, C, D, E, F, G$ or $H$ to any other, the sum of the flow rates is the same and does not depend on the path shape (taking into account the sign, if we move against the flow, the flow rate is subtracted).

It turned out that the total flow rate of water entering node $A$ is $q_{0}$. Find the flow rate in each of the channels $D E, B C$, and $G F$.
Answer: $\frac{4}{7} q_{0}, \frac{2}{7} q_{0}, \frac{3}{7} q_{0}$

Solution: First, due to symmetry, note that the flow in channel $\mathrm{BC}$ is equal to the flow in channel CD (denote it as $x$), and the flow in channel BG is equal to the flow in channel GD (denote it as $y$).

Then the total flow along the path $\mathrm{BCD}$ is equal to the flow along the path $\mathrm{BGD}: 2 x=2 y$. Therefore, $x=y$.

In channel $\mathrm{DE}$, $2 x$ already flows.

Denote the flow in channel GF as $q$. The same flow in channel $\mathrm{FE}$.

The total flow along the path GFE is $2 q$, and along the path GDE is $3 x$. From this, we get that $x=\frac{2}{3} q$. Therefore, in channel $\mathrm{DE}$, $2 x=\frac{4}{3} q$ flows.

The water flow entering node $\mathrm{E}$ (and, therefore, in A) is the total $q+\frac{4}{3} q=\frac{7}{3} q$. According to the problem statement, this is $q_{0}$. Therefore, $q=\frac{3}{7} q_{0}$, and $x=\frac{2}{7} q_{0}$.

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