Larry has Identified a point F, such that EF is parallel to AB. That is the key to one solution. Everything else is labeling points and creating lines between points, and working with properties of triangles. (It's harder than that makes it sound ;-) )
Here is another point that could be used for a different solution. I'll use "M" to make sure not to confuse with the solution Larry is following.
M is a point on BC such that MAB is 20*. From there, it is all about identifying isosceles and equilateral triangles, and line segments of the same length.