Tl;Dr : Prove that line passing through a vertex of triangle and intersection of diagonals of trapezoid formed by base and a parallel line divides both segments into two equal parts.
I was reading about Jewish(or Coffin) problems, (Link: https://arxiv.org/abs/1110.1556). I am not able to understand the solution of problem 16. I am putting the problem and the solution here.
You are given two parallel segments. Using a straightedge, divide one of them into six equal parts.
Given a segment and a parallel line we can always divide the segment into two equal parts. Take two points on the parallel line. Together with the ends of the segment the points form a trapezoid. Continue the sides to form a triangle. The segment passing through the third vertex of the triangle and the intersection of the diagonals of the trapezoid divides both parallel segments into two equal parts. Using the division method above, divide one of the segments into eight equal parts. Pick six of these parts consecutively. Then perform a homothety mapping their union onto the other segment. The center of the homothety is the intersection of the other sides of the trapezoid formed by the six segments and the target segment.
The part I am unable to understand is:
The segment passing through the third vertex of the triangle and the intersection of the diagonals of the trapezoid divides both parallel segments into two equal parts.
I tried to prove this but am unable to. I don't even know where to begin. Please help me in proving this.