Problem: Let $ABCD$ be a parallelogram.
Point $E$ lies on segment $CD$ such that
\[ 2\angle AEB = \angle ADB+\angle ACB, \]
and point $F$ lies on segment $BC$ such that
\[ 2\angle DFA = \angle DCA + \angle DBA. \]
Let $K$ be the circumcenter of triangle $ABD$. Prove that $KE=KF$.

Solution: Let the circle through $A$, $B$, and $E$ intersect $CD$ again at $E'$,
and let the circle through $D$, $A$, and $F$ intersect $BC$ again at $F'$.
Now $ABEE'$ and $DAF'F$ are cyclic quadrilaterals with two parallel sides,
so they are isosceles trapezoids.
From $KA=KB$, it now follows that $KE=KE'$,
whereas from $KA=KD$ it follows that $KF=KF'$.

Next, let the circle through $A$, $B$, and $E$ intersect $AC$ again at $S$.
Then
\[ \angle ASB = \angle AEB = \frac12(\angle ADB + \angle ACB)
  = \frac12(\angle ADB + \angle DAC) = \frac12 \angle AMB, \]
where $M$ is the intersection of $AC$ and $BD$.
From $\angle ASB = \frac12 \angle AMB$, it follows that $MS=MB$,
so $S$ is the point on $MC$ such that $MS=MB=MD$.
By symmetry, the circle through $A$, $D$, and $F$ also passes through $S$,
and it follows that the line $AS$ is the radical axis
of the circles $(ABE)$ and $(ADF)$.

By power of a point, we now obtain
\[ CE \cdot CE' = CS \cdot CA = CF \cdot CF', \]
from which it follows that $E$, $F$, $E'$, and $F'$ are concyclic.
The segments $EE'$ and $FF'$ are not parallel, so their
perpendicular bisectors only meet at one point, which is $K$.
Hence $KE=KF$.