Let S be a semicircle with diameter AB. The point C lies on the diameter AB and points E and D lie on the arc BA, with E between B and D. Let $S$ be a semicircle with diameter $AB$.  The point $C$ lies on the diameter $AB$ and points $E$ and $D$ lie on the arc $BA$, with $E$ between $B$ and $D$.  Let the tangents to $S$ at $D$ and $E$ meets at $F$.  Suppose that $\angle ACD = \angle ECB$.  Prove that $\angle EFD = \angle ACD + \angle ECB$.  
 A: I will assume that $C$ is closer to $B$ than to $A$, the other case is symmetric. If you join $D$ and $E$ to the center $O$, the quadrilateral $OEFD$ is inscriptible and therefore $\angle DOE=\pi-\angle DFE$. . You want to prove that the quadrilateral $DOCE$ is inscriptible as well (the circumscribed circle will be the same as before) and therefore $\angle DCE=\angle DOE$, since they are both inscribed on the same chord. But clearly $\angle DCE=\pi-\angle ACD-\angle ECB$ and your assertion follows. 
To prove that the quadrilateral $DOCE$ is inscriptible you can just extend $DC$  to the other side of the circle (call $G$ the point of intersection) and observe that, by the given relation $\angle ACD=\angle ECB$ you have that $E$ and $G$ are symmetric with respect to the diameter $AB$. That symmetry implies that $\angle OEC=\angle CGO$. You also have that $\angle CGO=\angle ODC$ since the triangle $GOD$ is isosceles. This establishes that $\angle OEC=\angle ODC$ which proves that $OCED$ is inscriptible, hence the desired assertion.
Hope this helps.
