26 May '12 05:56

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 ?

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60.13N / 25.01E- Joined
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26 May '12 20:46

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:*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 ?**

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- Joined
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2014.05.0131 May '12 05:19

Three possibilities:*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 ?**

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.