Point m is marked inside the parallelogram. Prove that the sum of the areas of triangles ABM

Point m is marked inside the parallelogram. Prove that the sum of the areas of triangles ABM and CDM is equal to half the area of parallelogram ABCD.

From point M we will construct a perpendicular to the base of AD. Then the area of the triangle АМD will be equal to: Samd = АD * МН / 2.

Let’s also draw the perpendicular MK to the side BC, then Svcm = BC * MK / 2.

Since the opposite sides of the parallelogram are equal, then Svcm = AD * MK / 2.

The segment KН is the height of the parallelogram, then Savsd = AD * KН = AD * (MK + MН).

The sum of the areas of triangles АМD and ВСМ is equal to:

Samd + Svsm = (AD * MН / 2) + (AD * MK / 2) = AD * (MH + MK) / 2 = Savsd / 2, which was required to prove.