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.