> 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.