- Joined
- 26 Apr 03
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26 May 12
Originally posted by talzamirNo (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:
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 ?
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
31 May 12
Originally posted by talzamirThree possibilities:
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 ?
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.