In a circle, two parallel chords are given. Construct a third chord that divides the previous two into 3 equal segments.
Let $A_{1} B_{1}$ be a chord (see the figure) that is a solution to the problem.

![](https://cdn.mathpix.com/cropped/2024_05_02_c098c097ecf6abca59fcg-1.jpg?height=529&width=553&top_left_y=229&top_left_x=775)

Draw lines through the endpoints of the given chords $a$ and $b$ that are parallel to them. We know that the segments cut by four parallel lines from a transversal are in the same ratio as the distances between the corresponding parallel lines. Therefore, $A_{1} B_{1}$ is a solution to the problem if and only if the distances between two adjacent parallel lines among the four are equal.

Accordingly, the construction procedure is as follows: Let $a^{\prime}$ be the reflection of $a$ over $b$, and $b^{\prime}$ be the reflection of $b$ over $a$. Let $a^{\prime}$ intersect the circle at points $A_{1}$ and $A_{2}$, and $b^{\prime}$ intersect the circle at points $B_{1}$ and $B_{2}$ (see the figure). Then the chords $A_{1} B_{1}$, $A_{1} B_{2}$, $A_{2} B_{1}$, and $A_{2} B_{2}$ satisfy the requirements of the problem.

The number of solutions is 4, 2, 1, or 0 depending on whether both reflections intersect the circle; one intersects and the other is tangent; both are tangent; or at least one does not intersect the circle.