Taiwan IMO Selection Camp

Take a point $T$ on ray $AE$ such that $AT = AB$; then since $BC \parallel AE$, we get
$$\angle CBT = \angle ATB = \angle ABT,$$
so $BT$ is the angle bisector of $\angle ABC$.

On the other hand, we have
$$ET = AT - AE = AB - AE = BC,$$
therefore quadrilateral $BCTE$ is a parallelogram, and $M$, the midpoint of diagonal $CE$, is also the midpoint of diagonal $BT$.

Next, let $K$ be the reflection of $D$ with respect to $M$. Then $OM$ perpendicularly bisects segment $DK$, hence $OD = OK$, meaning that point $K$ lies on the circumcircle of $\triangle BCD$. Thus $\angle BDC = \angle BKC$.

On the other hand, angle $BKC$ and angle $TDE$ are symmetric with respect to the point $M$, so $\angle TDE = \angle BKC = \angle BDC$. Therefore
$$\begin{aligned} \angle BDT &= \angle BDE + \angle EDT = \angle BDE + \angle BDC \\ &= \angle CDE = \angle ABC = 180^\circ - \angle BAT. \end{aligned}$$
This means that $A, B, D, T$ are four concyclic points, from which we obtain
$$\angle ADB = \angle ATB = \frac{1}{2} \angle ABC = \frac{1}{2} \angle CDE \text{, as required!}$$