forgive me

Very simple solution, but took me some time to realize. Notice, by simple angle chasing, that $\triangle ABD\stackrel{+}{\sim}\triangle AEC$ thus $\triangle AID\stackrel{+}{\sim}\triangle AJC$. Therefore $A$ is the Miquel point of $\{DI, CJ, IJ, BC\}$. Thus if $X=IJ\cap BC$, then $AIDX$ are concyclic. Hence $\angle AXD = 180^{\circ} - \angle AID = 90^{\circ} - \tfrac{\angle B}{2}$, which is fixed. Hence $X$ is the desired fixed point.

Here is a short solution with spiral similarity/linearity. Clearly $\triangle ABD \stackrel{+}{\sim} \triangle AEC$ so $\triangle AID \stackrel{+}{\sim} \triangle AJC$, and also $\angle AID = \angle AJC = 90^\circ + \tfrac12\angle ABC$. Let $K$ be the point on line $\overrightarrow{BC}$ such that $\angle AKB = 90^\circ - \tfrac12\angle ABC$. We claim that $K$ is the fixed point. Let $X$ be the point on $\overrightarrow{BC}$ satisfying $\angle ADI = \angle ACJ = \angle AXK$. Then $\triangle AID \stackrel{+}{\sim} \triangle AJC \stackrel{+}{\sim} \triangle AKX$. Since $D$, $C$, $X$ are collinear, it follows that $I$, $J$, $K$ are collinear, as wanted.

Let $BC$ and $IJ$ meet at $K$. Extend rays $AI$, $AD$, $AJ$, $BI$ to meet the circumcircle again at $P$, $F$, $N$, $M$. We use Pascal's theorem to eliminate $I$: \[ BMEPAN \overset{\text{Pascal}}{\implies} \overline{IJK} \parallel \overline{BN} \parallel \overline{PE}. \][asy][asy] size(11cm); pair A = dir(50); pair C = dir(-40); pair B = dir(180+40); pair E = dir(-40-45); pair F = dir(180+40+45); pair M = dir(midpoint(A--C)); pair D = intersectionpoint(A--F, B--C); pair I = incenter(A,B,D); pair J = incenter(A,C,E); pair K = intersectionpoint(I+10*(J-I)--I, B+10*(C-B)--B); pair N = circumcenter(J,C,E); pair P = B*C/N; pair L = extension(A,N,C,C+I-J); filldraw(A--B--D--cycle, invisible, deepgreen); filldraw(A--C--E--cycle, invisible, deepgreen); draw(F--A--E--cycle, lightblue); draw(P--A--N--cycle, lightblue); draw(B--M--E, lightblue); draw(D--K, deepgreen); draw(E--P, red); draw(B--N, red); draw(I--K, red); draw(C--L, red+dashed); dot("$A$", A, A); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, SW); dot("$E$", E, E); dot("$F$", F, F); dot("$P$", P, P); dot("$N$", N, N); dot("$M$", M, M); dot("$I$", I, dir(135)); dot("$J$", J, dir(120)); dot("$K$", K, dir(-90)); dot("$L$", L, dir(45)); filldraw(unitcircle, invisible, blue); [/asy][/asy] Define $L$ on line $AN$ such that $CL$ is parallel to these three lines. Using the classical fact that $JN = CN$ we are done upon noting that \[ \frac{BK}{BC} = \frac{JN}{LN} = \frac{CN}{LN} = \frac{\sin \angle NLC}{\sin \angle LCN} = \frac{\sin \angle ANB}{\sin \angle BNC} = \frac{AB}{BC}. \] Remark: In fact, this implies $KA = KB$. This opens the possibility of several other solutions. For example, $AJCK$ is cyclic, as is $AIDK$, and thus angle-chasing based solutions are possible.

Let $IJ$ meets $BC$ at $X$. We claim that $X$ is independent of $D$ by showing that the length $BX=BA$. Let $M$ be the midpoint of arc $AC$ not containing $B$ then by fact 5 $M,I,B$ are collinear, $M,J,E$ are collinear. Meanwhile from spiral similarity $$\triangle AIJ\sim\triangle ABE$$hence $\angle IAJ=\angle BAE=\angle BME=\angle IMJ$ which implies that $A,I,J,M$ are concyclic. Now since $MJ=MA$ by fact 5, $MI$ is the angle bisector of $\angle AIX$ again by fact 5. Since $BI$ is the angle bisector of $\angle ABX$, $A$ and $X$ are symmetric about $BI$ which implies $BX=BA$ as desired.

Notice that by the inscribed angle theorem $\angle ABD=\angle AEC$ and from the given $\angle BAD = \angle EAC$. Hence $\triangle BDA \sim \triangle ECA$. Consequently $\triangle DIA \sim \triangle CJA$. Then $\frac{AI}{AJ}=\frac{AD}{AC}$. Also, since $\angle CAJ + \angle IAD=\angle EAC$, we have that $\angle IAJ = \angle DAC$. Then $\triangle DAC \sim \triangle IAJ$, whence $\angle AIJ = \angle ADC$. Let $F=BC \cap IJ$. From the latter we have that quadrilateral $FDIA$ is cyclic. Since $\angle DIA=90^{\circ} + \frac{1}{2}\angle DBA$, we have that $\angle DFA=90^{\circ}-\frac{1}{2}\angle FBA$. Then $FB=BA$, and line $IJ$ always passes through the point $F$ on $BC$ such that $\triangle FBA$ is isosceles, which surely doesn't depend on $D$. $\square$

How do we consider degenerate cases?I mean when D=B,D=C.I think they give that the desired point is on BC.But how do we draw the incenter of ABD which is a single line?

a_bc wrote: How do we consider degenerate cases?I mean when D=B,D=C.I think they give that the desired point is on BC.But how do we draw the incenter of ABD which is a single line? In the actual contest it was specified that $D$ was different from $B$ and $C$(at least in the Spanish version).

Yes but considering degenerate cases gives us the hint where the point might be but here i dont understand how to use the degenerate case,why does the point have to be on BC in degenerate case?

Let $P$ be the midpoint of minor arc $AC$, and consider the circle with center $P$ and radius $PA=PC$. By the incenter-excenter lemma, $J$ lies on this circle. Let $X$ be the intersection of this circle with $BC$. Since $AB>BC$, this point lies outside the segment $BC$ (easy exercise.) This lets us dodge config issues. We claim that $X$ is the fixed point, and we'll do so with an angle chase that shows that $I,J,X$ are collinear. Observe that $APJI$ is cyclic by an easy angle chase on $\angle JAI=\angle BAE =\angle PBE = \angle IJP$. Thence, $\angle AJI = \angle API=\angle APB = \angle ACB$, and $\angle AJX = \angle ABX =\pi - \angle ACB$, so $\angle AJI + \angle AJX =\pi$, as desired.

Let $IJ$ meets $BC$ at $X$, $BI$ meets $EJ$ at $F$. Note that $F$ is always the midpoint of the minor arc $AC$. By angles chasing, we obtain that $AIJF$ is cyclic, moreover, we have $AF=FJ$ by incenter lemma. Hence, $\angle AIF=\angle JIF$. This implies $\triangle AIB \cong \triangle XIB$ and, consequently, $AJCX$ is cyclic. Note that $F$ is the circumcenter of $AJC$ and the radius of $(AJC)$ is independent from $D$. Hence, $X$ is independent from $D$. Remark The assumption $AB>BC$ can be omitted.

Because $\angle ABD = \angle AEC$ and $\angle BAD = \angle EAC$ we have that $A$ is the spiral center sending $EJC \to BID$. Thus the point $P = IJ \cap CD$ satisfies $AIDP$ concyclic. Since $\angle AID = \tfrac{1}{2} \angle B + 90^\circ$ is fixed we have that $\angle APD$ must also be fixed, and thus $P$ is fixed as desired.

$BI$ and $EJ$ meet at the midpoint of arc $AC$ of $(ABC)$ not containing $B$, say $M$. Easy angle chasing shows that $\measuredangle BAI=\measuredangle JAC$. Let $G$ be the reflection of $A$ over $BI$. Clearly $G$ lies on $BC$ and $MG=MA$ by symmetry. Hence, $MA=MJ=MC=MG$ by incenter-excenter lemma implying that $(AJCG)$ is cyclic. $$\measuredangle JGC=\measuredangle JAC=\measuredangle BAI=\measuredangle IGB=\measuredangle IGC$$So, $I, J, G$ are collinear, and moreover $G$ is independent of $D$.

From Fact 5 we conclude that $M = BI \cup EJ$ lies on $\odot (ABEC)$. Note that $BM$ is fixed, since $\triangle ABC$ is fixed. Let $T$ be the refelction of $A$ over $BM$. I claimed that $T$ is the desired fixed point. Claim. $T \in BC.$

Claim. $A$, $J$, $C$, $T$ are concyclic.

Claim. $M$, $A$, $I$, $J$ are concyclic.

Finally, $$\measuredangle AJT = \measuredangle ACT = \measuredangle ACB = \measuredangle AMB = \measuredangle AMI = \measuredangle AJI$$Therefore, $IJ$ always pass through $T$, which is fixed. $\blacksquare$

Let $IJ$ intersect $DC$ at $F$ Note that $A$ is the center of the spiral sending $BD$ to $EC$, so it is the center of the spiral also sending $ID$ to $JC$ Since $\angle{AIF}=\angle{ADF}$ it implies $A,I,D,F$ cyclic and similarly $A,J,C,F$ cyclic. $$\angle{AFB}=\angle{AFI}+\angle{IFD}=\angle{ADI}+\angle{IAD}=90-\frac{1}{2}\angle{ABC}$$ Hence, $F$ is independent of $D$

Let $\measuredangle$ denote directed angles modulo $180^\circ$. Let $M_B$ be the arc midpoint opposite $B$, and let the circle centered at $M_B$ through $C$ intersect $\overline{BC}$ again at $P \ne C$. Notice that $ABD \sim AEC$, so $\angle AIB=\angle AJE$ implies that $AIJM_B$ is cyclic. By the incenter-excenter lemma, $AJPC$ is cyclic, so we have $\measuredangle AJI=\measuredangle AM_BI=\measuredangle ACB=\measuredangle AJP$, which means $\overline{IJ}$ passes through $P$, as desired. $\square$

We prove a stronger statement, by allowing $D$ to vary along line $\overline{BC}.$ Let $K$ be the incenter of $\triangle ABC.$ Let $F=\overline{IJ} \cap \overline{BC}.$ Fix $\triangle ABC$ and animate $D$ projectively on $\overline{BC}.$ The map $D \mapsto I$ is projective by rotation about $B,$ and in particular, since $I$ varies on $\overline{BK},$ we have $\deg(I)=1.$ By the definition of $J,$ we have \[\angle AJC = 90^\circ + \frac{1}{2} \angle AEC = 90^\circ + \frac{1}{2}\angle B = \angle AKC.\]Thus, $J \in (AKC).$ In particular, by reflection about the internal angle bisector of $\angle A,$ the map $I \mapsto J$ is projective. It follows $\deg(J)=2.$ By Zack's Lemma, $\deg(\overline{IJ})=1+2-1,$ where $-1$ comes from $D=C,$ at which $I$ and $J$ coincide. By Zack's Lemma again, $\deg(F)=2-1=1,$ where the $-1$ here comes from $D=B,$ at which $\overline{IJ}$ and $\overline{BC}$ coincide. Thus, it suffices to check for two positions of $D,$ that $F$ is fixed (we cannot use $D=B$ or $D=C$ since we already used such cases to reduce degrees with Zack's Lemma). In particular, check for $D=\overline{AK} \cap \overline{BC}$ and $D=P_\infty$ that $F=(AKC) \cap \overline{BC}.$ Remark. After writing this, I realize perhaps a more natural way of writing this is to instead take initially $F := (AKC) \cap \overline{BC},$ and compute the degree of the hypothetical collinearity $\overline{IJF}.$ This avoids the use of Zack's Lemma, and with the above projective maps, along with the fact that here, we define $F$ to be fixed, the collinearity has degree $1+2=3,$ which would imply we need to verify it for four positions of $D.$ Here, we can take $D=B, D=C,$ along with the two aforementioned points.