albusgks

2022-04-30

If a line L separates a parallelogram into two regions of equal areas, then L contains the point of intersection of the diagonals of the parallelogram.

The figure shows a line L horizontally through the sides of the parallelogram.

This creates two trapezoids and I can intuitively show that if the bases of the trapezoids are not congruent then the areas can not be equal.

I just can not currently see how to relate the intersection with the diagonals.

Any help will be appreciated.

The figure shows a line L horizontally through the sides of the parallelogram.

This creates two trapezoids and I can intuitively show that if the bases of the trapezoids are not congruent then the areas can not be equal.

I just can not currently see how to relate the intersection with the diagonals.

Any help will be appreciated.

Diya Bass

Beginner2022-05-01Added 20 answers

draw in one of the diagonals, $d$. This will form two triangles using parts of $d$, $L$, and the sides of the parallelogram. Using properties of parallel lines, these triangles must be similar. Moreover, your discovery that the trapezoid bases must be equal implies that these triangles are actually congruent, and therefore $L$ intersects the diagonal at its midpoint (also the point of intersection with the other diagonal).

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