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.)
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.
A rotation about $180^\circ$ around the center of the parallelogram interchanges the two parts.
Hint: use angles to show that the two shapes [obtained by slicing the parallelogram] are similar, and then show that they are congruent.