I can sketch a proof based on assuming this "finite" result:

A). For any pentagonal star one of the 5 triangles will have area strictly smaller than that of the central pentagon. (I think a brute force attack should yield a proof here.) star http://img268.imageshack.us/img268/6733/proofay.png

The proof of the original problem would then go as follows.

b). A) generalizes to n-agons by considering the pentagon spanned by any 5 vertices. hexagon http://img263.imageshack.us/img263/1743/proofbe.png

c). b) implies that a tiling with polygons of EQUAL AREA is not possible unless all polygons are either triangles or quadrilaterals.

d). Take one 4-tile and continue tiling next to it inside the cone enclosed by the converging lines of 2 opposite edges; we have a sequence of quadrilaterals which must end with a triangle were the the 2 lines meet. This shows that the tiling must contain a 3-tile somewhere.

4-tile http://img202.imageshack.us/img202/7531/proofdm.png

e). By d) take a 3-tile and continue tiling outwards, inside each of the 3 beams generated by the lines of each pair of edges; the original 3-tile will be the first tile in each beam, but every other tile after it must be a 4-tile (build them one at a time and keep using c)). We can ignore what happens in the 3 leftover cones radiating from the 3 vertices.

3-tile http://img191.imageshack.us/img191/4290/proofen.png

f). In one of the 3 beams (which now look like ladders) take any one of the new rungs from step e) and extend it - that line will then collide with one of the other 2 beams (but cannot overlap with any of its rungs). That will cut one of the 4-tiles, creating a 5-tile.

contradiction! http://img585.imageshack.us/img585/4230/prooffh.png

Apologies for bumping up the question repeatedly while trying to edit my answer.