Can anyone solve this?

That's my solution during the contest. It is different from the official one. In order to have $C'K \parallel BB'$ o $ABB'C$ cyclic we need that $B'$ and $C'$ are both beyond $BC$ (with respect to $ABC$). Now $C'K \parallel BB' \iff JK \cdot JB' = JB \cdot JC'$ and $ABB'C$ is cyclic $\iff AJ \cdot JB' = BJ \cdot CJ$. So we need to prove that $ JK \cdot JB' = JB \cdot JC' \iff AJ \cdot JB' = BJ \cdot CJ $. Now $JK=CK-CJ= \frac{AC \cdot BC}{AC+AB} - BC+JB = JB- \frac{AB \cdot BC}{AC+AB}$ and $ JB' = AB'-AJ = AB-AJ$ and $JC'=AC-AJ$. So we need $$JB(AB^2-AC^2)=AB \cdot BC \cdot (AB-AJ) \iff AB \cdot AJ = AJ^2 + BJ \cdot CJ $$ or $$ BJ \cdot AC^2 + CJ \cdot AB^2 = AB \cdot BC \cdot AJ \iff AB \cdot BC \cdot AJ = AJ^2 \cdot BC + BJ \cdot CJ \cdot BC$$ But according to Stewart theorem we have $AB^2 \cdot CJ + AC^2 \cdot BJ = AJ^2 \cdot BC + BJ \cdot CJ \cdot BC$. And so we finally have that $C'K \parallel BB' \iff ABB'C$ is cyclic.

Let $C'' \in AB $ such that $AC''=AC;L =(ABC)\cap AK$ .we know that $AK.AL=AB.AC \implies KLB'C'$ is cyclic and $KLBC''$ is cyclic . If $B'\in ( ABC)$ then applying Reim to $ (KLB'C'),(ALB') [=ABC]$ yields $ KC'\parallel t $ which is the tangent to $(ABC)$ at $A$. let $A' $ point of $t$ in the same semi-plane bounded by $AK$ as $C$ .$\widehat{AB'B}=\widehat{ABB'}=\widehat{A'AB'}=\widehat{KC'B'} $ thus $KC'\parallel BB'$ conversly if $ KC'\parallel BB'$ then $KC'\parallel C'C'' \implies K,C',C'' $ are colinear . $KLBC''$ cyclic $ \implies \widehat{ABL}=\widehat{AKC''} =\widehat{AKC'}$ from $KLB'C' $ cyclic we deduce $ \widehat{ABL}= \widehat{C'B'L}= \widehat{AB'L}$ thus$B'\in (ABL)=(ABC)$

remark trisectrix is superfluous just cevian suffices

Let $ABC$ a triangle, let $K$ be the foot of the bisector relative to $BC$ and $J$ be the foot of the trisectrix relative to $BC$ closer to the side $AC$ ($3\angle JAC)=\angle CAB$ ). Let $C'$ and $B'$ be two point on the line $AJ$ on the side of $J$ with respect to $A$, such that $AC'=AC$ and $AB=AB'$. Prove that $ABB'C$ is cyclic if and only if lines $C'K$ and $BB'$ are parallel.