*Originally posted by THUDandBLUNDER*

**The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees.
**

The point D lies on AC such that angle DBC = 60 degrees.

The point E lies on AB such that angle ECB = 50 degrees.

Find angle CED.

I get that angle CED must be 80:

Without going through every step, I solved the following set of equations simultaneously. The equations follow from the Law of Sines and from the problem statement:

AD/[sin(130 - CED)] = AE/[sin(30 + CED)];

AE/sin(30) = AC/sin(130);

AD/sin(20) = AB/sin(140);

AB = AC;

Solving this set of equations, I get CED = 80.