Solve Geometry Problem: Find Angle EDC Without Cheating

[Jameson]

Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE.

Find what angle EDC is by using geometry only and no trigonometry.

Don't cheat and use protractors/rulers/all that stuff either! Go by pure geometric reasoning

My reasoning:

This is a tedious drawing so if you are going to procede I'll thank you in advance. Here is my reasoning for solving this problem.

Call the intersection of BE and CD point M. Let ADE=x, EDM=y, DEA=z, and DEM=140-z.

Thus we have the following system of equations:

(x+y)+20+10=180
x+20+z=180
50+y+(140-z)=180

This gives me a determinant of zero so this system although I think it is true doesn't tell me anything. Any ideas?

Thanks

 

[AKG]

Draw a line from E to CB that is parallel to AB, and label F as the point of intersection of this line with CB. Then use the stuff under the heading "Parallel Lines" on this page.