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 flow rate in the channel $A H$ is $q_{0}$. Find the flow rate in the channel $A B$, in the channel $B C$, and the total flow rate of water entering node $A$.
Answer: $\frac{4}{3} q_{0}, \frac{2}{3} q_{0}, \frac{7}{3} q_{0}$

Solution: First, due to symmetry, note that the flow in channel HG is also equal to $q_{0}$. Let the flow in channel $\mathrm{BC}$ and its symmetric channel $\mathrm{CD}$ be $x$, and the flow in channel $\mathrm{BG}$ and its symmetric channel GD be $y$.

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

The sum of the flows along the path AHG is $2 q_{0}$, and along the path $\mathrm{ABG}$ it is $3 x$. From this, we get that $x=\frac{2}{3} q_{0}$

The water flow entering node A is distributed into channels AB and AH. Therefore, the total water flow is $q_{0}+\frac{4}{3} q_{0}=\frac{7}{3} q_{0}$

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