Dear Mathlinkers, For a synthetic proof, you can see http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure I p. 41 fir the cocyclicity p. 17 for the nature of the point F Sincerely Jean-Louis

Let $S,T$ be the intersections of $CJ,CI$ with $\omega$, and let $SE,TE$ intersect $AK$ at $U,V$. Then $SA=SK=SJ=SD$ and $TB=TI=TK=TA$. Since $\triangle ATK$ is isosceles at vertex $T$, $AV/VK = AE/ED$. Similarly $AU/UK = AE/BE$. So $U,V;A,K$ are harmonic. Projecting from $E$ onto $\omega$ yields that $S,T;F,A$ are also harmonic. Thus $FT/AT =FS/AS$, which yields that $FT/TI=FS/SJ$. Hence $\triangle FTI \sim \triangle FSJ \Rightarrow \triangle FTS \sim \triangle FIJ \Rightarrow \angle IFJ =\angle TFS= \angle ICJ$.

Attachments:
czech.pdf (468kb)

see problem g8/imo short list 1999. by that we can solve this problem.

Let $T=EK\cap (ADB)$. It is well known that $T$ is the $A-$mixtillinear incircle of $\triangle ADB$. Let $P=CJ\cap (ABD), Q=CI\cap (ADB)$. And let $F=AK\cap (ADB)$. Clearly $EF$ is the diameter of $(ADB)$ and perpendicular to $DB$. So \[ -1=(E,F;D,B)\stackrel{K}{=}(T,A;Q,P)\implies AP\cdot TQ=TP\cdot AQ \]By fact 5 we know $AP=JP, AQ=IQ$. So, \[ \frac{AP}{TP}=\frac{AQ}{TQ}\implies \frac{JP}{TP}=\frac{IQ}{TQ} \]And We know $\angle TPJ=\angle TQI$ which means $\triangle TJP\sim\triangle TIQ$. This implies $\angle TJC=\angle TIC$.