Third of a Triangle

Is it possible to divide any triangle ABC into three parts by choosing points D and E on the side BC so that the angles CAD, DAE and EAB are all equal, and that the lengths BD = DE = EC = BC / 3 ?

Originally posted by talzamir
Is it possible to divide any triangle ABC into three parts by choosing points D and E on the side BC so that the angles CAD, DAE and EAB are all equal, and that the lengths BD = DE = EC = BC / 3 ?

No (if I understand the question correctly), the positions D and E can be found by trisecting D&E, which can be done with compass and straight-edge:
http://www41.homepage.villanova.edu/robert.styer/trisecting%20segment/

If the angles were also equal then BAC would have been trisected with compass and straight edge, which is impossible.
http://en.wikipedia.org/wiki/Angle_trisection

Originally posted by talzamir
Is it possible to divide any triangle ABC into three parts by choosing points D and E on the side BC so that the angles CAD, DAE and EAB are all equal, and that the lengths BD = DE = EC = BC / 3 ?

Three possibilities:

1) BD < BE
But then CAD > DAE. (DAE is a 'sub-angle' of CAD)

2) BD > BE
But then BD > ED (ED is part of the segment BD)

3) BD = BE
But then DE = 0

So, no.