If EQ is parallel to DF and AF intersects Q, it is easy to see that EQ: DF = 3: 4, EQ: AB = 1: 6, EG: GB = 1: 6, then the area of triangle AEG is 1/7 of triangle AEB, and the area of triangle AEB is 3 of parallelogram ABCD. Then it is calculated that the area of the triangular ADF is 1/9 of the total area, and the area of the shaded part is 1/9-3/56 = 29/504, 256 * 29/506 = 928/63 of the total area. The answer is horrible! Did I miscalculate?