Proving result in inscribed triangles. ABC is a triangle inscribed in a circle, and E is the mid-point of the arc subtended by BC different from the arc A on which A lies. If through E a diameter ED is drawn, show that $$\angle DEA=\frac{1}{2} \cdot \mid \angle ABC-\angle ACB \mid$$.
Additionally I denoted O as the circumcentre of the triangle, and F the point of intersection of ED and BC. I found that $ ED \perp BC $ and that $\mid BF \mid = \mid CF\mid $, and that $\angle EAD =90 $ degrees. But I haven't made any substantial progress. Can anyone help me?
Comes from 1988 IrMO.

 A: The observations you made, combined with the additional observation that $AE$ is the angle bisector of $\angle BAC$, pretty much solve the problem.
We will consider the case where $\angle ABC> \angle ACB$, and so $D$ lies on arc $AC$ not containing $B$, the case where $\angle ACB < \angle ABC$ is analogous, just with $D$ on arc $AB$ not containing $C$, (notice that if $\angle ABC=\angle ACB$, the problem is degenerate, and $D$ coincides with point $A$, so $DEA=0^{\circ}$). Let $M$ be the midpoint of $BC$. Let $AC\cap DE=P$. Since $\angle DMC$ is right, $\angle CPM=90^{\circ}-\angle ACB$. By vertical angles, $\angle APD=\angle CPM=90^{\circ}-\angle ACB$. Now as you noticed, because $DE$ is a diameter, $\angle DAE$ is right. But since $E$ is the midpoint of arc $BC$ not containing $A$, $AE$ is the angle bisector of $\angle BAC$, this is because as Jack D'Aurizio mentions: congruent arcs are subtended by congruent angles. So, $\angle DAP=90^{\circ}-\frac{\angle BAC}{2}$. Using triangle $ADP$, we get $\angle ADP=180^{\circ}-(90^{\circ}-\frac{\angle BAC}{2})-90^{\circ}-\angle ACB=\frac{\angle BAC}{2}+\angle ACB$. Now from right triangle $ADE$, we get $\angle DEA=180^{\circ}-90^{\circ}-(\frac{\angle BAC}{2}+\angle ACB)=90^{\circ}-\frac{\angle BAC}{2}-\angle ACB$. But from the original triangle $ABC$, we have $\angle ACB+\angle BAC+\angle ABC=180^{\circ}$, so $\frac{\angle BAC}{2}+\angle ACB=90^{\circ}+\frac{\angle{ACB}}{2}-\frac{\angle ABC}{2}$, so $\angle DEA=90^{\circ}-(90^{\circ}+\frac{\angle{ACB}}{2}-\frac{\angle ABC}{2})=\frac{\angle ABC}{2}-\frac{\angle ACB}{2}$
A: Assume that $A$ is to the right of $D$, as in the figure.Then
$$\eqalign{2\angle DEA&=\angle DCA+\angle DBA=(\angle ACB-\angle DCB)+(\angle DBC-\angle ABC)\cr &=\angle ACB-\angle ABC\ ,\cr}$$
since the triangle $\triangle(BDC)$ is isosceles at $D$.
