The figure you describe can certainly decompose into
six congruent (but possibly mirror flipped) triangular parts.
If the trapezoid has parallel sides of 2 and 6, and the non-parallel sides are 3, then it will easily divide into four congruent triangles (see below).
I'm not sure what you mean by "regular", but it's clear that if the two sides of a trapezoid are equal then it is symmetrical about a central altitude.
So if you construct two altitudes down from the upper vertices, you get a rectangle with congruent triangles on either side. That rectangle can be bisected along a diagonal to yield two more triangles identical to the first two -- if the dimensions of the trapezoid are 6-3-2-3 (or indeed 6-x-2-x, for any x>2)
If the dimensions are 6-x-3-x (x>1.5) then the rectangle will decompose into
four triangles identical to the first two, i.e., six parts in all.
Still, I can't say that it's impossible to solve the problem you stated -- it might simply involve pieces whose shape we haven't considered. (It wouldn't surprise me if it
were impossible, but the compelling logic of mathematical proof eludes me.
)
Can you re-check the wording of the problem?
