In article <firstname.lastname@example.org>, rb <email@example.com> wrote:
> > rb wrote : > >> Kaba : > >>> rb : > >>>> > >>>> http://img180.imageshack.us/img180/8286/20080218181526nq7.gif > >>>> > > > > Seems the drawing is distorted (y scale factor != x scale factor) > > > >>>> The angle to find is ECD. > >>>> The matter is solving it without trigonometric functions. > >>> > >>> Interesting problem, couldn't solve it either. I get 4 unknowns but > >>> only 3 linearly independent equations. > >> > >> I'm not sure if an equation system is needed. > > > > you may consider 180° = x + y + 20° as an equation ;-) > > > >> I guess drawing a few parallels and looking for properties like > >> angle bisector theorem, Thales' theorem, etc... should do the trick > >> by showing two similar triangles. > > > > Or maybe isosceles triangles, or even equilateral ones... > > > > Knowing the answer (ECD=20°) shows ABC is isosceles (AB=BC). > > But how to proove it directly ??? > > We already have another isosceles triangle if we extend line ED and > > BC to their intersect point F : FEB is isosceles (angles FEB = FBE) > > How to use this FEB triangle to get other properties ? > > Yes, I was working mainly on this triangle and on a parallel to BC > passing through D. This gives two 20° angles and also shows the > bisector theorem in the triangle EBD but I don't know if it's > necessary... > > > The problem here is we have no hint for which point/line would give > > a specific (hence right, isosceles, equilateral) triangle which > > would give the searched angles. > > > > This is similar to another problem, being discussed two weeks ago > > on a french newsgroup. > > I've put it on my Web site, as well as this (unsolved) one. > > > > <http://chephip.free.fr/pbm_en/pb226.html> > > > > The hint there was to add a point at a specific 20° angle, which > > results then into isosceles triangles. > > Nice one! > It's really very similar. I wonder if mine is a particular case too. > > > Without the hint, we need hard trigonometric calculations. > > (not very hard but still harder than "angles sum up to 180° in a > > triangle") > > These trig equations is the ones that lack to Kaba to solve his system. > > > > Unfortunately, I didn't either find a suitable point for this new > > problem. (the 20° trick works only in the other problem's angle values) > > > > Regards. > > Thanks for your help ;) > > I lost my notes :) but see http://chephip.free.fr/pbm_en/sol226.html the last figure looks the same as your problem in disguise.