Consider Δ ABD

We know that diagonals in a parallelogram bisect each other

∴ E is the midpoint of BD, AE is median of Δ ABD

∴ Area (Δ ADE) = Area (Δ AEB) (∵ Median divides the triangle into two triangles of equal areas)

Similarly we can prove

Area (Δ ADE) = Area (Δ DEC)

Area (Δ DEC) = Area (Δ CEB)

Area (Δ CEB) = Area (Δ AEB)

∴ Diagonals of a ‖ gm divide into four triangles of equal area.