Triangle CBD is a right triangle, so it can be shown that BC = 13 by the Pythagorean Theorem.
Since AC is a diameter, angle ABC is 90 degrees. From this it can then be shown that triangles ABC, BDC, and ADB are similar.
Using the ratios of the sides of similar triangles, the lengths of line segments AB and AD may be calculated.
One now knows the diameter and radius of the circle, and EO, OD, and ED can be calculated.
Knowing ED and DB, EB may be found by the Pythagorean Theorem.
The three sides of triangle ABE are now known. With this information, all the angles of triangle ABE can be found, using the Law of Cosines.
Add line segment OF to the drawing to create triangle FEO. Angle FEO equals angle AEB (vertical angles). Angle EOF is two times angle ABF (central angle and inscribed angle subtending the same arc).
Once you have two angles of a triangle, the third angle is easily found. One now has all the angles and one side (EO) of triangle FEO. Segment EF can now be found by Law of Sines and/or Cosines.
QED
PS I’m sure there are some easier ways to do this.
