If from any point in a diagonal of a parallelogram lines are be drawn to the opposite angles the parallelogram will be divided into two pairs of equivalent triangles.
The problem appeared in a periodical publication Mathematical Visitor and then included into a recent collection under the same name. It took the broadcast power of internet for the problem to reach overseas and a generalization emerge. Floor van Lamoen, The Netherlands, has noticed that point M does not have to lie on a diagonal of the parallelogram for there to be two pairs of triangles with equal sum of areas.
So let M be any point inside parallelogram ABCD. Triangles AMB and CMD have equal bases: AB =
. The sum of their altitudes ME + MF =
is fixed and does not depend on point M. We have
Area(AMB) + Area(CMD)
= AB·ME/2 + CD·MF/2
= AB·ME/2 + AB·MF/2
= AB·(ME + MF)/2
= AB·EF/2
=
.
It follows that the remaining area, Area(BMC) + Area(AMD) is also Area(ABCD)/2. So that the two are equal.
