A bit trivial for China TST. Note three facts: (i) There exists a circle tangent to rays $AB,BC,DC$ and $AD$, with center $L$ right on the angle bisectors $AE,BJ$ and $CK$. (ii) $I,J,K,B,C$ are concyclic. (iii) $\triangle FCE \sim \triangle ACL$. The main part is to prove $AL\parallel PK$ and $KP/KC=LA/LC$ (via computation), which show that $A,C,P$ are collinear. So $\triangle FCE$ and $\triangle PCK$ are also similar, and so are $\triangle PCF$ and $\triangle KCE$. Hence $\measuredangle CFP = \measuredangle CEK$, as desired.

GrosseFuge339 wrote: A bit trivial for China TST. Note three facts: (i) There exists a circle tangent to rays $AB,BC,DC$ and $AD$, with center $L$ right on the angle bisectors $AE,BJ$ and $CK$. (ii) $I,J,K,B,C$ are concyclic. (iii) $\triangle FCE \sim \triangle ACL$. The main part is to prove $AL\parallel PK$ and $KP/KC=LA/LC$ (via computation), which show that $A,C,P$ are collinear. So $\triangle FCE$ and $\triangle PCK$ are also similar, and so are $\triangle PCF$ and $\triangle KCE$. Hence $\measuredangle CFP = \measuredangle CEK$, as desired. The fact (1) isn't true.. It needs to be $AB+CD=BC+DA$, not $AB+BC=AD+DC$. I think there is a typo in the question.

What a strange problem. First we interpret the condition \(AB+BC=AD+DC\). By the ``excircle'' version of Pitot (see IMO 2008/6), there is a circle \(\omega\)—say, centered at \(X\)—tangent to line \(BC\) and the extensions of rays \(AB\), \(DC\), \(AD\). Therefore, lines \(BJ\), \(AE\), \(CK\) concur at \(X\). [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=lightred; pen sec=orange; pen tri=mediumblue; pen tri2=lightblue; pen qua=purple+pink; pen qui=pink+magenta; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair aa,bb,cc,dd,AA,B,C,D,O,A,I,K,J,X,EE,F,P; aa=dir(150); bb=dir(270); cc=dir(320); dd=-aa*cc/bb; AA=2/(1/dd+1/aa); B=2/(1/aa+1/bb); C=2/(1/bb+1/cc); D=2/(1/cc+1/dd); O=circumcenter(B,C,D); A=O+(B-O)*(D-O)/(AA-O); I=incenter(A,B,C); K=incenter(D,B,C); J=2*extension( (B+C)/2,O,A,I)-I; X=extension(B,J,C,K); EE=2*foot(O,A,X)-A; F=2O-A; P=extension(A,C,K,K+X-A); draw(B--X--K,qui);draw(X--A,tri2); filldraw(circumcircle(B,I,C),qfil,qua); filldraw(incircle(A,B,C),sfil,sec); filldraw(incircle(B,C,D),sfil,sec); draw(A--P,pri);draw(B--D,pri); filldraw(circumcircle(B,C,D),fil,pri); fill(A--B--C--D--cycle,fil); draw(B--A--D--C,pri); filldraw(I--B--C--cycle,tfil,tri); filldraw(P--K--J--cycle,tfil,tri); dot("\(A\)",A,unit(A-O)); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(250)); dot("\(D\)",D,unit(D-O)); dot("\(I\)",I,unit(I-J)); dot("\(J\)",J,dir(260)); dot("\(K\)",K,NE); dot("\(X\)",X,SE); dot("\(E\)",EE,dir(250)); dot("\(F\)",F,unit(F-O)); dot("\(P\)",P,E); [/asy][/asy] Claim: \(\overline{PK}\parallel\overline{AX}\). Proof. Observe the following: \begin{align*} \measuredangle PKC&=\measuredangle PKJ+\measuredangle JKC=\measuredangle IBC+\measuredangle JBC=90^\circ+2\measuredangle IBC=90^\circ+\measuredangle ABC,\\ \measuredangle AXC&=90^\circ+\measuredangle(\overline{AX},CE)=90^\circ+\measuredangle AEC=90^\circ+\measuredangle ABC. \end{align*}\(\blacksquare\) Claim: \(P\) lies on line \(AC\). (In particular, \(\triangle CKP\sim\triangle CXA\).) Proof. From the given similarity \(\triangle PKJ\sim\triangle IBC\) alongside \(\triangle XBC\sim\triangle XKJ\), we have \[KP=\frac{IB\cdot KJ}{BC}=\frac{IB\cdot XJ}{CX}.\]However, note that \(\measuredangle XJA=\measuredangle BJI\) and \(\measuredangle XAJ=\measuredangle EAB+\measuredangle BAI=\measuredangle KBC\), so \[\frac{XA}{XJ}=\frac{\sin\angle XJA}{\sin\angle XAJ}=\frac{\sin\angle BJI}{\sin\angle KBC}=\frac{IB}{CK},\]so \(IB\cdot XJ=CK\cdot XJ\), Combining these, we have \[KP=\frac{CK\cdot KJ}{CX},\]so \(\triangle CKP\sim\triangle CXA\) by SAS. It follows that \(A\), \(C\), \(P\) are collinear. \(\blacksquare\) Finally, \(\measuredangle CFE=\measuredangle CAE=\measuredangle CAX\) and \[\measuredangle ACX=90^\circ+\measuredangle ACE=90^\circ+\measuredangle AFE=\measuredangle FAE=\measuredangle FCE,\]so \(\triangle CEF\sim\triangle CXA\). Combining with the above claim, \(\triangle CEF\sim\triangle CKP\). Since spiral similarities come in pairs, \(\triangle CEK\sim\triangle CFP\), so \(\measuredangle CEK=\measuredangle CFP\). The desired conclusion follows.

Can someone please tell the motivation behind the above solution. Like, first of all the figure is too tough to draw by hand (even on GeoGebra). Secondly, there are so many points in the question and so many angles to chase, then how to keep doing going in the right direction ?

TheUltimate123 wrote: What a strange problem. First we interpret the condition \(AB+BC=AD+DC\). By the ``excircle'' version of Pitot (see IMO 2008/6), there is a circle \(\omega\)—say, centered at \(X\)—tangent to line \(BC\) and the extensions of rays \(AB\), \(DC\), \(AD\). Therefore, lines \(BJ\), \(AE\), \(CK\) concur at \(X\). [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=lightred; pen sec=orange; pen tri=mediumblue; pen tri2=lightblue; pen qua=purple+pink; pen qui=pink+magenta; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair aa,bb,cc,dd,AA,B,C,D,O,A,I,K,J,X,EE,F,P; aa=dir(150); bb=dir(270); cc=dir(320); dd=-aa*cc/bb; AA=2/(1/dd+1/aa); B=2/(1/aa+1/bb); C=2/(1/bb+1/cc); D=2/(1/cc+1/dd); O=circumcenter(B,C,D); A=O+(B-O)*(D-O)/(AA-O); I=incenter(A,B,C); K=incenter(D,B,C); J=2*extension( (B+C)/2,O,A,I)-I; X=extension(B,J,C,K); EE=2*foot(O,A,X)-A; F=2O-A; P=extension(A,C,K,K+X-A); draw(B--X--K,qui);draw(X--A,tri2); filldraw(circumcircle(B,I,C),qfil,qua); filldraw(incircle(A,B,C),sfil,sec); filldraw(incircle(B,C,D),sfil,sec); draw(A--P,pri);draw(B--D,pri); filldraw(circumcircle(B,C,D),fil,pri); fill(A--B--C--D--cycle,fil); draw(B--A--D--C,pri); filldraw(I--B--C--cycle,tfil,tri); filldraw(P--K--J--cycle,tfil,tri); dot("\(A\)",A,unit(A-O)); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(250)); dot("\(D\)",D,unit(D-O)); dot("\(I\)",I,unit(I-J)); dot("\(J\)",J,dir(260)); dot("\(K\)",K,NE); dot("\(X\)",X,SE); dot("\(E\)",EE,dir(250)); dot("\(F\)",F,unit(F-O)); dot("\(P\)",P,E); [/asy][/asy] Claim: \(\overline{PK}\parallel\overline{AX}\). Proof. Observe the following: \begin{align*} \measuredangle PKC&=\measuredangle PKJ+\measuredangle JKC=\measuredangle IBC+\measuredangle JBC=90^\circ+2\measuredangle IBC=90^\circ+\measuredangle ABC,\\ \measuredangle AXC&=90^\circ+\measuredangle(\overline{AX},CE)=90^\circ+\measuredangle AEC=90^\circ+\measuredangle ABC. \end{align*}\(\blacksquare\) Claim: \(P\) lies on line \(AC\). (In particular, \(\triangle CKP\sim\triangle CXA\).) Proof. From the given similarity \(\triangle PKJ\sim\triangle IBC\) alongside \(\triangle XBC\sim\triangle XKJ\), we have \[KP=\frac{IB\cdot KJ}{BC}=\frac{IB\cdot XJ}{CX}.\]However, note that \(\measuredangle XJA=\measuredangle BJI\) and \(\measuredangle XAJ=\measuredangle EAB+\measuredangle BAI=\measuredangle KBC\), so \[\frac{XA}{XJ}=\frac{\sin\angle XJA}{\sin\angle XAJ}=\frac{\sin\angle BJI}{\sin\angle KBC}=\frac{IB}{CK},\]so \(IB\cdot XJ=CK\cdot XJ\), Combining these, we have \[KP=\frac{CK\cdot KJ}{CX},\]so \(\triangle CKP\sim\triangle CXA\) by SAS. It follows that \(A\), \(C\), \(P\) are collinear. \(\blacksquare\) Finally, \(\measuredangle CFE=\measuredangle CAE=\measuredangle CAX\) and \[\measuredangle ACX=90^\circ+\measuredangle ACE=90^\circ+\measuredangle AFE=\measuredangle FAE=\measuredangle FCE,\]so \(\triangle CEF\sim\triangle CXA\). Combining with the above claim, \(\triangle CEF\sim\triangle CKP\). Since spiral similarities come in pairs, \(\triangle CEK\sim\triangle CFP\), so \(\measuredangle CEK=\measuredangle CFP\). The desired conclusion follows. Um isn't $ABCD$ circumscribed about $\gamma$