- 08 Jun '04 06:17 / 1 editBest I can tell, it is not solvable, because I have been able to plug in any angle to cde and the resulting triangles on the diagram all equal 180. If I am wrong, message me why (don't spoil it).

This is what I found: Where X is the intersection between BE and CD, you can figure out all of other angles from several of the resulting triangles with no stress (Triangles ACD, AED, and XBC). But, plug in any number for angle CDE (some number between 1 and 109, since angle DXE is 70), and the resulting angles (BED, BEA, and ADE) are forced into place, and all angles work equally well (resulting in triangles with 180 degrees). - 08 Jun '04 18:13i leave out degrees, and with Z(ABC) i mean angle ABC, with T(ABC) i mean triangle ABC.

I'm still doubting if it is solvable. I've gotten this far:

70 < Z(CDE) =< 110

0 =< Z(BED) < 40

30 =< Z(ADE) <70

90 < Z(AED) =< 130

Z(CDE) + Z(BED) = 110

Z(ADE) + Z(AED) = 160

Z(ADE) + Z(CDE) = 140

Z(AED) + Z(BED) = 130

For wich i used that AB=AC, CB=CE, DA=DC (follows from equal angles at the bases of T(CEB) and T(DCA)).

This is as far as i can get, though i already know that, for example

Z(CDE) cannot be 110, for then Z(BED) would be 0 and thus D=B.

Furthermore i've tried something with the projection of D on AC, for is exatly halfway AC...