(2013 Nanjing Consortium Second Model) As shown in the figure, in ABCD, E is a point on the diagonal BD, and the line segments FG and HP passing through the point E intersect the four sides of the parallelogram respectively.

Answer: Solution: When FG∥BC and HP∥AB, the relationship between the areas of the two shaded parts in the figure is uniquely determined.

Reason: ∵ Quadrilateral ABCD is a parallelogram,

∴∠A=∠C, AB∥CD, AD∥BC,

∵AB∥PH, EF∥AD,

∴AD∥FG∥BC, AB∥HG∥CD,

∴∠ADB=∠DBC, the quadrilateral EFBP and the quadrilateral DHEG are parallelograms,

∵In △DAB and △BCD

∠A＝∠C∠ADB＝∠DBBCD＝BD,

∴△DAB≌△BCD (AAS),< /p>

∴S△DAB=S△BCD,

Similarly, S△DHE=S△DEG, S△EFB=S△BPE,

∵S quadrilateral AFEH=S△ABD-S△BFE-S△DHE, S quadrilateral EPCG=S△BCD-S△BEP-S△DEG,

∴S quadrilateral AFEH=S quadrilateral EPCG

< p>∴When FG∥BC and HP∥AB, the size relationship between the areas of the two shaded parts in the figure is uniquely determined.

So choose: C.