Proof that a line cuts in half the area of a parallelogram iff it goes through the intersection of the diagonals?

Question

I read a theorem in a book which says that a line bisects a parallelogram iff it goes through the intersection of the diagonals. The edge case of this is of course if the line is one of the diagonal which creates 2 triangle. It is well known that these 2 triangle have the same area. Is the full theorem true? Can you give me an elementary proof, for this? (I am still in secondary school.)

Answer 1

Step 1. First observe that indeed, if a line goes through the intersection of the diagonals, then bisects the parallelogram into two figures of equal area.

Step 2. If a line $C$ does not go through the intersection of the diagonals, then design the one which is parallel to $C$ and goes through the intersection of the diagonals. This is going to help you understand why "iff" in the statement you want to prove.

Answer 2

Hint: use angles to show that the two shapes [obtained by slicing the parallelogram] are similar, and then show that they are congruent.

Answer 3

A rotation about $180^\circ$ around the center of the parallelogram interchanges the two parts.