Let $$ABCD$$ be a parallelogram. Let $$E$$ and $$F$$ be midpoints of $$AB$$ and $$BC$$ respectively. The lines $$EC$$ and $$FD$$ intersect in $$P$$ and form four triangles $$APB$$, $$BPC$$, $$CPD$$ and $$DPA$$. If the area of the parallelogram is 100 sq. units, what is the maximum area in sq. units of a triangle among these four triangles?

Ratios of areas are unchanged by an affine map, so take the unit square with $$A = (0,0)$$, $$B = (1,0)$$, $$C = (1,1)$$, $$D = (0,1)$$, giving $$E = \left(\frac{1}{2}, 0\right)$$ and $$F = \left(1, \frac{1}{2}\right)$$. Solving $$y = 2x-1$$ with $$y = 1 - \frac{x}{2}$$ gives $$P = (0.8, 0.6)$$, and the four triangles then have areas $$0.3, 0.1, 0.2, 0.4$$ of the whole. The largest is $$0.4 \times 100 = 40$$ square units.