Berkeley Math Circle

Solution:
Let $E$ lie on segment $CD$ so that $CE = BC$. Then, $DE = DC - CE = AD$. So, $\triangle BCE$ and $\triangle ADE$ are both isosceles. Let $\alpha = \angle DAE = \angle AED$, and let $\beta = \angle CEB = \angle EBC$. Then $\angle CDA = \angle EDA = \pi - 2\alpha$, but since the quadrilateral is cyclic, this gives $\angle ABC = 2\alpha$. Likewise, $\angle BCD = \pi - 2\beta$ and $\angle DAB = 2\beta$.

Now, let the circumcircle of triangle $ABE$ intersect line $CD$ again at $F$. We will assume that $F$ lies on the same side of $E$ as $D$ (the other case is completely analogous). Since quadrilateral $ABEF$ is cyclic, we have $\angle ABF = \angle AEF = \alpha = \angle ABC / 2$, and $\angle FAB = \pi - \angle BEF = \angle CEB = \beta = \angle DAB / 2$. So $F$ lies on the bisectors of angles $DAB$ and $ABC$, which means it is precisely the intersection point specified in the problem statement. Since $F$ was constructed to lie on $CD$, we are done.