My solution involved breaking it into right angle triangles as per the attached, then assigning an arbitrary length to the bottom side of one of the lower triangles in the attached picture.

1) Determine missing angles to add up to 180 per triangle: EBD=20, ECD=10,CBD=40,BEC=30, intersection angles of 50 and 130.
2) Create 2 right-angle triangles from the 60-70-50 lower triangle and solve for h: h=xtan(60)
3) Determine f: sin(70)=h/f, therefore by substitution of h, f=xtan(60)/sin(70)
4) Determine c: sin(10)=c/f, therefore by substitution of f, c=sin(10).xtan(60)/sin(70)
5) Determine g: sin(60)=h/g, therefore by substitution of h, g=xtan(60)/sin(60)
6) Determine b: sin(20)=b/g, therefore by substitution of g, b=sin(20).xtan(60)/sin(60)

Now solve for lengths e and d
7) Determine e: sin(40)=c/e, therefore by substitution of c, e=sin(10).xtan(60)/sin(70)/sin(40)
8) Determine d: sin(30)=b/d, therefore by substitution of b, d=sin(20).xtan(60)/sin(60)/sin(30)

9) Determine a: Using the law of cosines: a^2 = e^2 d^2 - 2e.d.cos(50), therefore by substitution of e and d, a = sqrt< (sin(10).xtan(60)/sin(70)/sin(40))^2 (sin(20).xtan(60)/sin(60)/sin(30))^2 - 2.cos(50).(sin(10).xtan(60)/sin(70)/sin(40)).(sin(20).xtan(60)/sin(60)/sin(30)) >

10)Determine angle CED: Using the law of sines: sin(CED)/e = sin(50)/a, therefore by substitution of e and a,

angle CED = arcsin {sin(50).sin(10).xtan(60)/sin(70)/sin(40) / sqrt <(sin(10).xtan(60)/sin(70)/sin(40))^2 (sin(20).xtan(60)/sin(60)/sin(30))^2 - 2.cos(50).(sin(10).xtan(60)/sin(70)/sin(40)).(sin(20).xtan(60)/sin(60)/sin(30)) >}

angle CED = 20 degrees