WRONG Let the parallel line through $E $ to the bisector of angle $B $ intersect $AC $ at $K $ ,and define $T $ analogously. Simple angle-chasing yields that $AEKI $ and $BFTJ $ is cyclic. So there are two cases: 1)$I,J,K $ and $T $ are collinear; or 2) $IK\parallel JT $ (which can be gotten easily) The first case is just what problem wants from us.So we will look at the second case: First chase some angles to get $C,F,K $ and $J $ are concyclic.Now define new point $K'\equiv JT\cap AC $.Again chase some angles to get $C,F,K'$ and $J $ are concyclic.But since $C,K,K'$ are collinear $\implies $ $K\equiv K'$. So done...

tenplusten wrote: Let the parallel line through $E $ to the bisector of angle $B $ intersect $AC $ at $K $ ,and define $T $ analogously. I think this $K $ should be distinguished from the one in the Question. Btw the defination of $T $ is sort of vague. Can you elucidate it?

According to this topic: https://artofproblemsolving.com/community/u57217h1280308p6732388 https://artofproblemsolving.com/community/c6h1208052p5974754 It is easy to see that $IJ $ is perpendicular to bisector of angle $APB$. Then by some simple angle chasing we are done.

Let $EI,FJ$ intersects $\odot(APD),\odot(BCP)$ at $Q,R$ respectively, and $S$ be the intersection of $EI$ and $FJ.$ Notice that $\angle SEF=\tfrac{1}{2}\angle AED=\tfrac{1}{2}\angle APD=\tfrac{1}{2}\angle CPB=\tfrac{1}{2}\angle BFC=\angle SFE,$ which implies that $S$ lies on the perpendicular bisector of $EF.$ Meanwhile we have $\angle PEF=\angle PDA=\angle PCB=\angle PFE,$ which follows that $SP$ is the angle bisector of $\angle QSR.$ Claim: $QR$ is parallel to $IJ.$ Proof: We first show that $P,Q,R$ are collinear, this is because $\angle APQ+\angle APB+\angle BPR=\tfrac{1}{2}\angle APD+\tfrac{1}{2}\angle BPC+\angle APB=180^\circ.$ Notice that $\triangle AQP\sim \triangle BRP.$ Now that $SP$ bisects $\angle QSR,$ we get \[\frac{SQ}{SR}=\frac{PQ}{PR}=\frac{QA}{RB}=\frac{QI}{RJ}.\]Hence $IJ$ is parallel to $QR$ as desired. $\square$ In conclusion we have $\angle EIJ=\angle EQP=\angle EAP,$ so $A,I,K,E$ are concyclic. [asy][asy] size(7cm); pointpen=black; pathpen=black; defaultpen(fontsize(9pt)); pair A,B,C,D,P,I,J,Q,R,E,F,S; A=dir(-160); B=dir(-20); C=dir(30); D=dir(110); P=IP(A--C,B--D); E=OP(circumcircle(A,P,D),A--B); F=IP(circumcircle(B,P,C),A--B); I=incenter(A,D,E); J=incenter(C,F,B); Q=OP(L(E,I,5,5),circumcircle(A,P,D)); R=OP(L(F,J,5,5),circumcircle(B,P,C)); S=extension(I,E,J,F); filldraw(anglemark(A,Q,P,3),magenta); filldraw(anglemark(A,D,P,3),magenta); filldraw(anglemark(A,C,B,3),magenta); filldraw(anglemark(P,R,B,3),magenta); D(unitcircle); D(circumcircle(A,P,D),blue); D(circumcircle(B,P,C),blue); D(A--B--C--D--cycle); D(A--C); D(B--D); D(D--E); D(C--F); D(I--J,red+linewidth(1.2)); D(Q--S--R, dashed); D(Q--R,red+linewidth(1.2)); D(Q--A, dashed); D(R--B,dashed); D(E--P--F,deepgreen+linewidth(1.1)); D(S--P,dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$P$",P,dir(85)*2); dot("$E$",E,dir(-90)); dot("$F$",F,dir(-65)); dot("$I$",I,dir(60)); dot("$J$",J,dir(-90)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); [/asy][/asy]

$\frac{QA}{RB}=\frac{QI}{RJ}$ could you explain this ?

Can anyone solve this problem by inversion?

Let $M,N$ be the circumcenters of $\triangle AID$ and $\triangle BJC$. Let $S=\overline{IE} \cap \overline{JF}$. Note that $\overline{MN}$ is the bisector of angle $APD$. Moreover, $\angle PMA=\angle PDA=\angle PCB=\angle PNB$. Hence $\triangle PAM \sim \triangle PBN$. Consequently, \begin{align*} \frac{MI}{NJ} = \frac{MA}{NB} = \frac{PA}{PB}= \frac{\sin PBA}{\sin PAB} = \frac{\sin SNM}{\sin SMN} = \frac{SM}{SN}\end{align*}proving $\overline{IJ} \parallel \overline{MN}$. Now it is $2\angle (IJ, AC)=\angle APD=2\angle AEI$ proving that $A,I,K,E$ are concyclic. $\blacksquare$

Let $X$ and $Y$ be the midpoints of shorter arcs $AD$ and $BC$ in the circles $\circ ADPE$ and $\circ BCPF$ respectively. $Z$ is the intersection point of $XE$ and $YF$. It is obvious that $X, P, Y$ are collinear. Claim 1: $\triangle AXP$ and $\triangle BYP$ are similar. Proof: $\angle AXP = \angle ADP = \angle PCB = \angle BYP$ and $\angle APX = \frac {\angle APD}{2} = \frac {\angle BPC}{2} = \angle BPY$ Claim 2: $XY||IJ$ Proof: $\frac{XI}{YJ} = \frac{XA}{YB} = \frac{PA}{PB} = \frac{\sin \angle PBA}{\sin \angle PAB} = \frac{\sin \angle PYZ}{\sin \angle PXZ} = \frac{XY}{YZ}$ Now we get back to our problem. $\angle IKA = \angle XPA = \frac {\angle DPA}{2} = \frac {\angle DEA}{2} = \angle IEA$ $Q.E.D.$

Attachments:

Note that \[ \angle BJC=90^{\circ}+ \frac{\angle BFC}{2} = 90^{\circ}+ \frac{\angle BPC}{2} =90^{\circ}+ \frac{\angle DPA}{2}=90^{\circ}+ \frac{\angle DEA}{2}= \angle DIA. \] Applying Trig Ceva on $BCJ$ wrt $A$ and $DAI$ wrt $B$ we get \[ \frac{\sin \angle AJB}{\sin \angle CJA} = \frac{\sin \angle AIB}{\sin \angle BID}. \] The last couple of points imply $\angle AJB= \angle AIB$ and thus $AIJB$ is cyclic. Finally note that \[ \angle JBA+ \angle AIJ=180^{\circ} \implies \angle EIK = \angle EAK. \]We conclude $AIKE$ is cyclic.

Solved with kingu. We actually struggled with this quite a bit. Let $M$ and $N$ be the intersections of $\overline{EI}$ and $\overline{FJ}$ with $(APD)$ and $(BPC)$ respectively. Let $T= \overline{EI} \cap \overline{FJ}$. Now, we can get started. Claim : The points $M$,$N$ and $P$ are collinear. Proof : We note that, \[2\measuredangle MPD = 2\measuredangle MED = \measuredangle AED = \measuredangle APD = \measuredangle CPB = \measuredangle CFD = 2\measuredangle NFB = 2\measuredangle NPB\]This means that $\measuredangle MPD = \measuredangle NPB$ from which it is quite clear that $M-N-P$ as claimed. Now, we have the most important step of the solution. Claim : Lines $MN$ and $IJ$ are parallel. Proof : To see why this is true, we first show that $\triangle MAP \sim \triangle NBP$. This is beacause, \[\measuredangle APM \sim \measuredangle CPN = \measuredangle CFN = \measuredangle NFB = \measuredangle NPB\]and \[\measuredangle PMA = \measuredangle PDA = \measuredangle BDA = \measuredangle BCA = \measuredangle BCP = \measuredangle BNP \] Now, note that by the Incenter-Excenter Lemma, $MI=MA$ and $NJ=NB$. Thus, \[\frac{MI}{NJ} = \frac{MA}{NB} = \frac{AP}{PB} = \frac{\sin \angle ABP }{\sin \angle PAB}\]Further, \[\frac{TM}{TN}=\frac{\sin \angle TNM}{\sin \angle TMN}=\frac{\sin \angle FNP}{\sin \angle EMP}=\frac{\sin \angle FBP}{\sin \angle EAP}=\frac{\angle ABP}{\sin \angle PAB}\] From which it follows that, \[\frac{MI}{MJ}=\frac{TM}{TN}\]which proves the claim. Now, we simply note that, \[\measuredangle JIE = \measuredangle NME = \measuredangle PME = \measuredangle PAE = \measuredangle KAE\]which implies that points $A,I,K$ and $E$ are indeed concyclic which was the desired conclusion.

pretty easy. Rename $K$ to $L$ since I'm bad and let $K=EI\cap FJ$. Let $M$ and $N$ denote $EI\cap(ADE)$ and $FJ\cap(BCF)$ as arc midpoints. We angle chase $M,N,P$ collinear. Then, \[\frac{MK}{NK}=\frac{\sin MNK}{\sin NMK}=\frac{\sin ABP}{\sin BAP}=\frac{AP}{BP}\]by LoS. But as $(AMDP)\sim (BNCP)$ we get that $\frac{AP}{BP}=\frac{MA}{NB}=\frac{MI}{NJ}$, we get $\frac{MK}{NK}=\frac{MI}{NJ}$ so $IJ \parallel MN$. To finish, \[\measuredangle EIL=\measuredangle EIJ=\measuredangle EMN=\measuredangle EMP=\measuredangle EAP=\measuredangle EAL\]