Çözüldü Triangle Area in Paralellogram

In a parallelogram ABCD, if E and F are the midpoints of sides CD and BC, respectively, what is the ratio of the area of triangle AEF to the area of parallelogram ABCD?

Height of the ΔABF and ΔCEF passing through F: h
|AB| = 2a
|CE| = |CD| = a
Area(ΔCEF) = a·h / 2 = S ⇒ a·h = 2S
Area(ΔABF) = 2a·h / 2 = 2S
Area(ΔADE) = a·2h / 2 = 2S
Area(ABCD) = 2a·2h = 4(a·h) = 4·2S = 8S
As the area of the rectangle AFCE is the half of the area of the parallelogram, it is 8S / 2 = 4S
Therefore, the area of the triangle ΔAEF is 4S - Area(ΔCEF) = 4S - S = 3S
Finally, Area(ΔAEF) / Area(ABCD) = 3S / (8S) = 3 / 8.

Source (Turkish): "ÖSS Geometri", FEM Publications, October 2003, Page 243, Problem 11