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 ?
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.
The hint there was to add a point at a specific 20° angle, which results then into isosceles triangles. 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)