$AX,BC$ and the line through $I$ perp to $AI(l_1$) is concurrent. Similarly $BC, AY$ and the line through $J$ perp to $AJ(l_2)$ are concurrent. These concurrencies can be proved by radical axis theorem with circles $(ABC), (BICJ),(AXI),(AYJ)$. Now the conclusion follows from the fact that $l_1,l_2$ and $MK$ are parallel where $M=AI \cap (ABC)$ and M is the midpoint of $IJ$. EDIT: We have to prove that $XY \parallel BC$. If $M$ is the midpoint of arc $BC$ and $N$ is the midpoint of are $BAC$ then $-1=(P_\infty,M;I,J)=(N,M;X,Y)$

A Loong solution. WLOG assume $AB<AC$. Let $A'$ be the $A$-antipode on $(ABC)$ and $N$ be the midpoint of the minor arc $\widehat{BC}$. Since $\angle{A'XA}=\angle{AXI}=90^{\circ}$ and $\angle{A'YA}=\angle{JYA}=90^{\circ}$, $X, I, A'$ and $Y, A', J$ are collinear. Let $H$ be the orthocenter of $\triangle A'XY$ and $M$ be the midpoint of $\overline{XY}$. It suffices to proof that $H, M, K$ are collinear as $H$ is the reflection of $A$ across $M$ (which imply $A, M, H$ would be collinear). The main observation is that $K$ is actually the orthocenter of $\triangle IJA'$. As $\measuredangle{NCK}=\measuredangle{NCB}=\measuredangle{NAB}=\measuredangle{CAN}=\measuredangle{CA'N},$ $\triangle NCK \sim \triangle NA'C$. The observation then follows from $NK \perp IJ$ and $NA' \cdot NK=NC^2=IN^2$. By considering the complete quadrilateral $IXYJ$ and point $A'$ with the Gauss-Bodenmiller theorem, or by dropping altitudes and calculating powers of points, we can show that $K$, $H$ are on the radical axis of $(IY)$ and $(XJ)$. Let $D, E$ be the foot from $I, J$ to $XY$ respectively, then clearly the reflection of $D$ over $M$ is $E$ due to $IN=JN$. Finally, $D$ is on $(IY)$ and $E$ is on $(XJ)$ yields $$\text{Pow}_{(YI)}(M)=-DM \cdot MY=-XM \cdot ME=\text{Pow}_{(XJ)}(M),$$which gives that $M$ is also on the radical axis of $(IY)$ and $(XJ)$. Therefore, $A, M, K$ are collinear.

Here is my generalization for it. Generalization: Given a $\triangle ABC$ and two points $U$, $V$ such that $\triangle ABU\stackrel{+}{\sim}\triangle AVC$. $X$, $Y$ are two points lying on $\odot(ABC)$ such that $\angle AXU=\angle AYV=90^\circ$. $K\in BC$ satisfies $KU=KV$. Then if the pedal circle of $U$ with respect to $\triangle ABC$ is tangent to $BC$, $AK$ bisects $\overline{XY}$. In paticular, when $U$ is the incenter of $\triangle ABC$, the previous statement holds.

Let incircle touch $BC, CA, AB$ at $D, E, F$ respectively Let A-excircle touch $BC, CA, AB$ at $D_1, E_1, F_1$ respectively Then $IEAXF$ and $JF_1YAE_1$ are cyclic Let M be midpoint of minor arc $BC$ of circumcircle of $\Delta ABC$ By incenter-excenter lemma $MI=MJ$ and hence $KM\perp IJ$ We will prove that 1) $X, D, M$ collinear 2) $Y, D_1, M$ collinear 1) $\angle XCE=\angle XCA=\angle XBA=\angle XBF$ $\angle XFB=180-\angle XFA=180-\angle XEA=\angle XEC$ $\implies \Delta XFB\sim \Delta XEC$ $\implies \frac{XB}{XC}=\frac{BF}{CE}=\frac{BD}{CD}$ By converse of angle bisector theorem $\angle BXD=\angle CXD$ or equivalently $X, D, M$ collinear 2) $\angle CF_1Y=\angle AF_1Y=\angle AE_1Y=\angle BE_1Y$ $\angle YCF_1=180-\angle ACY=180-\angle ABY=\angle YBE_1$ $\implies \Delta CF_1Y\sim \Delta BYE_1$ $\implies \frac{BY}{CY}=\frac{BE_1}{CF_1}=\frac{BD_1}{CD_1}$ By converse of angle bisector theorem $\angle BYD_1=\angle CYD_1$ or equivalently $Y, D_1, M$ collinear We know that $BD=CD_1$ hence by symmetry $XY\parallel BC$ Let $N$ be midpoint of $XY$ Then $MN\perp XY$ Let circumcircles of $JF_1YAE_1$ and $\Delta MNY$ intersect at $L$ We will show that $A, N, L, K$ are collinear $\angle NLY=\angle NMY=90-\angle NYM=90-\angle MYX=90-\angle MAX=90-\angle MAY=90-\angle JAY=\angle AJY=\angle ALY$ $\implies A, N, L$ collinear We will show that $JMLK$ is cyclic $\angle IMK=90=\angle IDK \implies IDMK$ cyclic Also $NM\parallel ID$ $\angle IKM=180-\angle IDM=\angle DMN$ $\angle ALJ=90=\angle MLY \implies \angle ALY=\angle JLM$ $\angle JLM=\angle ALY=\angle NLY=\angle NMY=\angle NMX=\angle IKM=\angle JKM$ $\implies JKLM$ cyclic $\angle MLK=180-\angle MJK=180-\angle MYN=180-\angle NLM$ $\implies N, L, K$ collinear $\implies A, N, K$ collinear

Let $D$ be the $A-$ antipode of $(ABC)$.Let $L=AI \cap (ABC)$.Then due to the right angles , $X-I-D , Y-D-J , K-D-L$.Now Let $R=AI \cap BC$.And $P=DR \cap (ABC)$.Then $$-1=(A,R;I,J) \stackrel{D}=(A,P;X,Y)$$thus $AP$ is the $A-$ symmedian of $\triangle AXY$. Also $$\angle DKR =\frac{B-C}{2} =\angle RAD \implies (ARDK)$$.Also $$\angle AXD =\angle XDl =\angle JDL =\angle YDl $$Thus $AI$ is the bisector of $XAY$.This means that $BX=CY \implies XY \parallel BC$. Now , $$\angle PAL= \angle RDL =\angle LAK$$Thus $AK$ is the reflection of the $A-$ symmedian across $AI$ , the angle bisector , which means that $AK$ is the median , as desired.

FishHeadTail wrote: In acute triangle $ABC$ ($AB \neq AC$), $I$ is its incenter and $J$ is the $A$-excenter. $X, Y$ are on minor arcs $\widehat{AB}$ and $\widehat{AC}$ respectively such that $\angle{AXI}=\angle{AYJ}=90^{\circ}$. $K$ is on line $BC$ such that $KI=KJ$. Proof that line $AK$ bisects $\overline{XY}$. Let $A'$ be the $A$-antipode in $\triangle ABC$ and $D=\overline{AI} \cap \overline{BC}$. Move a point $P$ on line $\overline{AI}$ and a point $Q$ on $\odot(ABC)$ such that $Q$ lies on the circle with diameter $AP$. Note that $P \mapsto Q$ is a perspectivity at $A' \in \odot(ABC)$, hence is a projective mapping. Let $L=\odot(AD) \cap \odot(ABC) \setminus A$ and note that $(XY; LA)=(IJ; DA)=-1$ by the above mapping. Finally, the image of $L$ under a $\sqrt{AB \cdot AC}$ inversion at $A$ followed by reflection in $AI$ is the point $K$; hence $\overline{AK}$ is a median in $\triangle AXY$ as $\overline{AL}$ is a symmedian, proving the claim.

Here's unnecessarily complicated and long solution. Let $I,I_B,I_C$ be the incenter, $B$-excenter and $C$-excenter of $\triangle ABC$. Let $A_0$ be point on $(ABC)$ such that $\overline{AA_0}\parallel \overline{BC}$. Let $A'$ be the antipode of $A$ wrt $(ABC)$. Let $N,M$ be midpoint of arcs $BAC,BC$, respectively. Claim. $\overline{XY}\parallel \overline{BC}$. Proof. The desired is actually equivalent to show that $\measuredangle YA'N= \measuredangle NA'X$. Observe that $A'$ lies on the perpendicular bisector of $\overline{I_BI_C}$. By DDIT, $(\overline{A'I},\overline{A'X}),(\overline{A'I_B},\overline{A'I_C}),(\overline{A'B},\overline{A'C})$ are reciprocal pairs of some involution. Hence, we conclude the angle bisector of $\angle BA'C,\angle XA'Y,\angle I_BA'I_C$ coincide, as desired. $\square$ Let $D=\overline{AI}\cap\overline{BC}$, let $E=\overline{A'E}\cap (ABC)$, let $F=\overline{AK}\cap (ABC)$. Claim. $\overline{EF}\parallel \overline{BC}$. Proof. Define $f(\bullet)=\pm\frac{\bullet B}{\bullet C}$ with the choice of signs as usual. By ratio lemma, \begin{align*} f(E)=\frac{f(D)}{f(A')}=\frac{f(A)\cdot f(M)}{f(A')}=\frac{f(A)}{f(K)}=\frac{1}{f(F)}, \end{align*}which yields that $\overline{EF}\parallel \overline{BC}$. $\square$ Therefore, finally, \begin{align*} -1=(A,D;I,J)\overset{A'}{=}(A,E;X,Y)\overset{\infty}{=}(A_0,F;Y,X)\overset{A}{=}(\infty,\overline{XY}\cap\overline{AK};Y,X), \end{align*}which gives $\overline{AK}$ bisects $\overline{XY}$, as desired. $\blacksquare$

Solved with Anshu droid Let $AI \cap BC = D$ and let $M$ be the midpoint of $\overarc{BC}$ First we use $\sqrt{bc}$ inversion along with the reflection across the angle bisector. Note that this swaps $B \iff C$ and $I \iff J$ $X$ maps to $X'$ which lies on $BC$ such that $X'J \perp AJ$ and $Y$ maps to $Y'$ which also lies on $BC$ such that $Y'I \perp AI$. Notice that there is a homothety centred at $D$ taking $I$ to $J$ and $Y''$ to $X'$. This means $$\frac{IY'}{JX'}=\frac{ID}{JD}=\frac{AI}{AJ} \implies \triangle AIY' \sim \triangle AJX' \implies \angle Y'AI = \angle X'AI$$where the last step follows because of sine law on $\triangle BDJ$ and $\triangle BAJ$. This also implies $$\angle XAI = \angle YAI \implies A-X-Y' \text{ and } A-Y-X'$$and $$\frac{AX}{AY}=\frac{AY'}{AX'} \implies XY \parallel BC$$Now just notice that this homothety also takes $M$ to $K$, that is the midpoint of $IJ$ to $K$ which means that $K$ must be the midpoint of $X'Y'$ which by homothety at $A$ taking $\triangle AXY$ to $\triangle AY'X'$ means that $AK$ must bisect $XY$ as desired $\blacksquare$ Simplest solution in this thread involving ratios and homothety (similar triangles)

Anshu-Droid's solve again . Solved with BVKRB- , this was my first time g-solving, and it was really fun lol . My solution has the proof of the first part pretty small and easy, and finish similar with that of BVKRB-. CGMO 2021 P2 wrote: In acute triangle $ABC$ ($AB \neq AC$), $I$ is its incenter and $J$ is the $A$-excenter. $X, Y$ are on minor arcs $\widehat{AB}$ and $\widehat{AC}$ respectively such that $\angle{AXI}=\angle{AYJ}=90^{\circ}$. $K$ is on line $BC$ such that $KI=KJ$. Proof that line $AK$ bisects $\overline{XY}$. Let the perpendicular to $AI$ passing through $I$ intersect $BC$ at $Y'$, the perpendicular to $AJ$ passing through $J$ intersect $BC$ at $X'$, $F$ be the midpoint of the minor arc $\overarc{BC}$ of $\odot ABC$, $Y'' = BC \cap AX$, $X'' = BC \cap AY$. Now we start off by inverting at $A$ with radius $\sqrt{AB \cdot AC}$ followed by a reflection wrt the Angle Bisector of $\angle BAC$. So, $$B \xleftrightarrow{} C$$$$BC \xleftrightarrow{} \odot ABC$$$$Y \xleftrightarrow{} Y'$$$$X \xleftrightarrow{} X'$$ Now by Radical Center Theorem on $\{ \odot ABC, \odot AXI, \odot IBC \}$ we get that $Y''I$ is tangent to $\odot AXI \implies \angle AIY'' = 90^{\circ} \implies Y'' = Y'$ and similarly $X'' = X'$. So, $\overline{A - X - X'}$ and $\overline{A - Y - Y'}$ are collinear triples, so, $\angle IAX = \angle IAY \implies FX = FY$ and since $FB = FC$ we get that $XY \parallel BC$. Now since $KI = KJ$, so $K$ lies on the perpendicular bisector of $IJ$, we get $K$ is midpoint of $X'Y'$(simple angle and similar triangle chase). Thus the homothety centred at $A$ that maps $XY$ to $Y'X'$, also maps $AK \cap XY$ to $K$, thus $AK \cap XY$ is the midpoint of $XY$ and we are done .

Really good problem! Let $AX \cap BC=X', AY \cap BC=Y'$. By homothety ($XY \parallel BC$ by angle chasing), we want $K$ to be the midpoint of $X'Y'$. Add $AI \cap (ABC)=M$; by radical axis $IX', JY' \perp AI$ (say, by taking $(IJ), (AI), (ABC)$ for the first one and similarly for the second one) and in addition, $MK \perp AI$, so $IX' \parallel MK \parallel JY'$ and we are done since $M$ is midpoint of $IJ$.

Obv $A, I, J$ collinear, $K, A', N$ collinear. Then we have $XY||BC$. It's easy to prove: $IS \perp IJ$. Then $KS=KT$. $\square$

Let $M$ be the midpoint of $\overline{XY}$. By the Inscribed Angle Theorem, $\angle AXI = \angle AYJ = \angle AXJ$. Therefore, the angles $\angle AXI$ and $\angle AXJ$ are supplementary, so $\angle AXJ = 180^{\circ} - 90^{\circ} = 90^{\circ}$. [asy][asy] unitsize(1 cm); pair A, B, C, I, J, K, M, X, Y; A = (0,0); B = (15,0); C = (2.5,15); I = (5,5); J = extension(A, I, B, C); X = (10,0); Y = (2.5,10); M = (3.725,5); K = extension(I, J, X, Y); draw(A--B--C--cycle); draw(I--J); draw(X--Y); draw(A--K); label("$A$", A, S); label("$B$", B, S); label("$C$", C, NW); label("$I$", I, NE); label("$J$", J, NW); label("$X$", X, S); label("$Y$", Y, NE); label("$K$", K, NW); label("$M$", M, N); [/asy][/asy] Then by the Exterior Angle Theorem, $\angle AKJ = \angle BAC + \angle AXJ = \angle BAC + 90^{\circ}$. Since $I$ is the incenter of $\triangle ABC$, $IK$ is an angle bisector of $\angle BAC$. Therefore, $\angle AKI = \angle BAC/2$. Hence, $$\angle AKJ = \angle AKI + \angle IKJ = \frac{\angle BAC}{2} + \angle IKJ = \frac{\angle BAC}{2} + \angle BAC = \frac{3 \angle BAC}{2}.$$This means that $\angle AKJ = \angle AKI + \angle IKJ$ is an angle bisector of $\angle AKJ$, so $AK$ bisects $\angle AKJ$. Therefore, $AK$ bisects $\overline{XY}$.

Solved with erkosfobiladol Let $S$ be the antipode of $A$ in $(ABC)$. Let $M$ be the midpoint of $\widehat{BC}$. $K$ is the intersection of $MS$ and $BC$. $S,I,X$ are collinear. $J,S,Y$ are collinear. Let $JS\cap (BIC)=P$ and $SI\cap (BIC)=Q$. $K$ is the orthocenter of $ISJ$ hence $I,P,K$ and $J,Q,K$ are collinear. Take the inversion centered at $M$ with radius $MB$. $(ABC)$ goes to the line $BC$ thus $A^*,X^*,Y^*\in BC$. $K\leftrightarrow S, A\leftrightarrow A^*$ The line $YPSJ$ goes to $(MY^*PKJ)$ with diameter $JK$ hence $JY^*\perp BC$. The line $XISQ$ goes to $(MX^*QKI)$ with diameter $IK$ hence $IX^*\perp BC$. We get that the incircle is tangent to $BC$ at $X^*$ and $A-$excircle is tangent to $BC$ at $Y^*$. We have $MX^*=MY^*\iff MX=MY$ So $XY\parallel BC$. $D=AI\cap BC$. Let $E=X^*$ which is the altitude from $I$ to $BC$ and $F=Y^*$ which is the altitude from $J$ to $BC$. Let $Q$ be the midpoint of $BC$, $R$ be the altitude from $S$ to $BC$, $G=(ABC)\cap SR$. $I', J'=(BIC)\cap MG$ where $I'G<J'G$. Let $T$ be the midpoint of $XY$. $\angle GSC=\angle RSC=\angle C=\angle ACB$ thus we have $AG\parallel BC$. $$\left. \setlength{\arraycolsep}{2pt} \begin{array}{rcl} -1=(G,D';I',J')=(R,D';F,E) \\[4pt] QF^2=QD'.QR \\[4pt] Pow_Q(EMF)=Pow_Q(DMSR) \\[4pt] T^*\in (DMS) \\[4pt] \end{array}\right\} \implies A,T,K \text{ are collinear.}$$