Olimpiade Italiana di Matematica

Let us first set $\theta = \angle CAJ$, so that $\angle JAB = 2\theta$ and $\angle JAK = \theta / 2$.

If the quadrilateral $ABB'C$ is cyclic, then $\angle CBB' = \theta$, since it subtends the same arc as the angle $\angle CAB'$. On the other hand, since $\angle AB'B = \angle ABB' = (180^\circ - 2\theta)/2$ by construction, the angles $\angle JBB'$ and $\angle BB'J$ are complementary, so $AJ$ is orthogonal to $BC$.

Now, $\angle AC'C = \angle ACC' = (180^\circ - \theta)/2$, hence $\angle JCC' = 90^\circ - \angle AC'C = \theta / 2$. But then $\angle C'CK = \theta / 2 = \angle C'AK$; this implies that the quadrilateral $AKC'C$ is cyclic, and hence $\angle CKC' = \angle CAC' = \theta = \angle CBB'$. Since they form congruent corresponding angles with the transversal line $BC$, the lines $C'K$ and $BB'$ are therefore parallel.

Suppose now that the lines $C'K$ and $BB'$ are parallel. The corresponding angles $\angle AC'K$ and $\angle AB'B$ formed with line $AB'$ must be congruent, hence both equal to $(180^\circ - 2\theta)/2$ (since triangle $ABB'$ is isosceles). It follows that $\angle KC'C = \angle KC'A + \angle AC'C = 180^\circ - \frac{3}{2}\theta$; hence $\angle KC'C$ is supplementary to $\angle CAK$, and the quadrilateral $CAKC'$ is cyclic. But then $\angle B'BC = \angle C'KC = \angle C'AC = \theta$ (where the first equality follows from the parallelism and the second from the cyclicity of $CAKC'$); hence the quadrilateral $CABB'$ is also cyclic.