It is well known due to the Apollonius Circle theorem that, the vertices of triangle and feet of internal and external angle bisectors of the opposite angle form harmonic pair. Projecting through the excentre finishes the problem.

So cuttee uwu Sharygin P19 wrote: Let $I$ be the incenter of triangle $ABC$, and $K$ be the common point of $BC$ with the external bisector of angle $A$. The line $KI$ meets the external bisectors of angles $B$ and $C$ at points $X$ and $Y$ . Prove that $\angle BAX = \angle CAY$ We begin with defining $I_A$ as the $A$ Excentre. Define $D$ as $AI\cap BC.$ Note that $A-I-D-I_A$ and $\angle DAK=90^{\circ}.$ To prove that $\angle BAX=\angle CAY,$ it is enough to show that $\angle XAI=\angle YAI$ as $\angle BAI=\angle CAI.$ Since we have $90$ degree and angle bisectors, by Right angles and bisectors lemma, it's enough to show that $(K,I;X,Y)=-1.$ Projecting from $I_A$ on $BC.$ We get $(K,I;X,Y)=(K,D;B,C).$ But note that $(K,D;B,C)=-1$ as $\angle KAD=90^{\circ},\angle BAD=\angle CAD.$ So by Right angles and bisectors lemma, we get $(K,D;B,C)=-1.$

Easily done by Barycentric and DDIT ( or say Isogonal line lemma) -

jelena_ivanchic wrote: So cuttee uwu Sharygin P19 wrote: Let $I$ be the incenter of triangle $ABC$, and $K$ be the common point of $BC$ with the external bisector of angle $A$. The line $KI$ meets the external bisectors of angles $B$ and $C$ at points $X$ and $Y$ . Prove that $\angle BAX = \angle CAY$ We begin with defining $I_A$ as the $A$ Excentre. Define $D$ as $AI\cap BC.$ Note that $A-I-D-I_A$ and $\angle DAK=90^{\circ}.$ To prove that $\angle BAX=\angle CAY,$ it is enough to show that $\angle XAI=\angle YAI$ as $\angle BAI=\angle CAI.$ Since we have $90$ degree and angle bisectors, by Right angles and bisectors lemma, it's enough to show that $(K,I;X,Y)=-1.$ Projecting from $I_A$ on $BC.$ We get $(K,I;X,Y)=(K,D;B,C).$ But note that $(K,D;B,C)=-1$ as $\angle KAD=90^{\circ},\angle BAD=\angle CAD.$ So by Right angles and bisectors lemma, we get $(K,D;B,C)=-1.$ same! the 2nd favorite problem!

Let $I$ be the incenter of $\triangle ABC$, and $K$ be the common point of $BC$ with the external bisector of angle $A$. The line $KI$ meets the external bisectors of angles $B$ and $C$ at points $X$ and $Y$. Prove that $\angle BAX = \angle CAY$. $\textbf{Solution}$ We'll use the Angle Bisector theorem. If $AB=AC$, then the external angle bisector of $\angle BAC$ would be parallel to $BC$, so $K$ wouldn't exist. Now WLOG let $AB>AC$ and let $C_{1}=AB\cap KI$, $B_{1}=IK\cap AC$. Denote $XC_{1}=a$, $C_{1}I=b$, $IB_{1}=c$, $B_{1}Y=d$, $YK=e$. By the Angle bisector theorem for $\triangle C_{1}BK$, $\triangle B_{1}CK$, $\triangle C_{1}AB_{1}$ we have that: \begin{align} \frac{a}{a+b+c+d+e}=\frac{C_{1}C}{XK}=\frac{C_{1}B}{BK}=\frac{C_{1}I}{IK}=\frac{b}{c+d+e}\\ \frac{c}{c+d+e}=\frac{B_{1}I}{IK}=\frac{B_{1}C}{CK}=\frac{B_{1}Y}{YK}=\frac{d}{e}\\ \frac{d+e}{b+c+d+e}=\frac{B_{1}K}{KC_{1}}=\frac{AB_{1}}{AC_{1}}=\frac{B_{1}I}{IC_{1}}=\frac{c}{b} \end{align}From $(2)$ we have that: \[\frac{c}{c+d+e}=\frac{d}{e}\Longrightarrow ec=d(c+d)+ed\Longrightarrow \boxed{e=\frac{d(c+d)}{c-d}}\]From $(3)$ we have that: \[\frac{d+e}{b+c+d+e}=\frac{c}{b}\Longrightarrow b=\frac{c(c+d+e)}{d+e-c}=\frac{c\left(c+d+\frac{d(c+d)}{c-d}\right)}{d+\frac{d(c+d)}{c-d}-c}=\]\[=\frac{c((c+d)(c-d)+d(c+d))}{(d-c)(c-d)+d(c+d)}=\frac{c^2(c+d)}{c(3d-c)}=\frac{c(c+d)}{3d-c}\Longrightarrow \boxed{b=\frac{c(c+d)}{3d-c}}\]From $(1)$ we have that: \[\frac{a}{a+b+c+d+e}=\frac{b}{c+d+e}\Longrightarrow a=\frac{b(b+c+d+e)}{c+d+e-b}=\]\[=\frac{\frac{c(c+d)}{3d-c}\left(\frac{c(c+d)}{3d-c}+c+d+\frac{d(c+d)}{c-d}\right)}{c+d+\frac{d(c+d)}{c-d}-\frac{c(c+d)}{3d-c}}=\]\[=\frac{\frac{c(c+d)}{3d-c}\left(\frac{c}{3d-c}+1+\frac{d}{c-d}\right)}{1+\frac{d}{c-d}-\frac{c}{3d-c}}=\]\[=\frac{\frac{c(c+d)}{3d-c}\left(c(c-d)+(c-d)(3d-c)+d(3d-c)\right)}{(c-d)(3d-c)+d(3d-c)-c(c-d)}=\]\[=\frac{c(c+d)}{3d-c}\cdot\frac{2cd}{4cd-2c^2}=\frac{cd(c+d)}{(2d-c)(3d-c)}\Longrightarrow \boxed{a=\frac{cd(c+d)}{(2d-c)(3d-c)}}\]Now we will prove the following: $\textbf{Lemma}$ $\frac{XI}{IY}=\frac{XK}{KY}$ $\textit{Proof:}$ Using the formulas for $a,b,e$ in $c,d$ we have that:\[\frac{XI}{IY}=\frac{XK}{KY}\iff \frac{a+b}{c+d}=\frac{a+b+c+d+e}{e}\]\[\iff \frac{\frac{cd(c+d)}{(2d-c)(3d-c)}+\frac{c(c+d)}{3d-c}}{c+d}=\frac{\frac{cd(c+d)}{(2d-c)(3d-c)}+\frac{c(c+d)}{3d-c}+c+d+\frac{d(c+d)}{c-d}}{\frac{d(c+d)}{c-d}}\]\[\iff \frac{c}{3d-c}\left(\frac{d}{2d-c}+1\right)=\frac{\frac{cd}{(2d-c)(3d-c)}+\frac{c(2d-c)}{(3d-c)(2d-c)}+\frac{(c-d)+d}{c-d}}{\frac{d}{c-d}}\]\[\iff \frac{c}{3d-c}\cdot\frac{3d-c}{2d-c}=\frac{(3cd-c^2)(c-d)+c(3d-c)(2d-c)}{d(3d-c)(2d-c)}\]\[\iff c=\frac{(3cd-c^2)(c-d)+c(3d-c)(2d-c)}{d(3d-c)}\]\[\iff c=\frac{3cd^2-c^2d}{d(3d-c)}\]\[\iff 3d^2c-c^2d=3cd^2-c^2d\]The last line is obviously true, so, going back the equivalences we have proven the Lemma.\newline\newline Let $X'\in IK$, so that $\angle X'AI=\angle YAI$ and $X'\not\equiv Y$. Then since $AI$ is the angle bisector of $\angle X'AY$ and $\angle IAK=\frac{1}{2}(\angle BAC+180^{\circ}-\angle BAC)=90^{\circ}$, therefore $AK$ is the external angle bisector of $\angle X'AY$ and we can use the angle bisector theorem for $\triangle X'AY$ we have that: $$\frac{X'I}{IY}=\frac{AX'}{AY}=\frac{X'K}{KY}\Longrightarrow \frac{X'I}{X'K}=\frac{IY}{KY}$$However, by the Lemma we know that $\frac{XI}{XK}=\frac{IY}{KY}$. Let $t=KX'-KX$, where $t$ is not necessarily positive, it's just a real number. We have that (obviously $I$ lies between $K$ and $X$ on $\overline{KX}$): \[\frac{XI}{XK}\overset{\textbf{Lemma}}{=}\frac{IY}{KY}=\frac{X'I}{X'K}\]\[\Longrightarrow \frac{XI}{XK}=\frac{X'I}{X'K}=\frac{XI+t}{XK+t}\Longrightarrow XI\cdot t=XK\cdot t\Longrightarrow t(XK-XI)=0\]However, obviously $XI\neq XK$, so $t=0$, which means that $X\equiv X'$, so, by the construction of $X'$ it follows that $\angle XAI=\angle YAI$, however $\angle BAI=\angle CAI$ as $I$ is the incenter of $\triangle ABC$, so: $$\angle BAX=\angle XAI-\angle BAI=\angle YAI-\angle CAI=\angle CAY$$Which is what we wanted to prove, so we've solved the problem.

Let $L=AI\cap BC.$ Clearly $Z=XB\cap CY$ is $A-\text{excenter},$ so $(XYIK)=(BCLK)=-1.$ Since $AI\perp AK,$ line $AI$ bisects angle $XAY$ as desired.

Let lines $AB$ and $AC$ intersect line $KI$ at $E$ and $F$, respectively. [asy][asy] size(12cm); pair A = dir(160), B = dir(230), C = dir(-50), I = incenter(A, B, C), K = extension(A, rotate(-90, A)*I, B, C), X = extension(B, rotate(90, B)*I, K, I), Y = extension(C, rotate(-90, C)*I, K, I), E = extension(A, B, K, I), F = extension(A, C, K, I), D = extension(A, I, B, C); draw(A--K--B--A--C--B, blue); draw(K--Y^^A--D, royalblue); draw(X--A--Y--D--X, deepcyan); draw(X--B--I--C--Y, mediumblue); string[] names = {"$A$", "$B$", "$C$", "$D$", "$E$", "$F$", "$I$", "$K$", "$X$", "$Y$"}; pair[] points = {A, B, C, D, E, F, I, K, X, Y}; pair[] ll = {dir(100), B, C, D, dir(60), dir(80), dir(65), K, dir(140), dir(70)}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Since $\angle XBI = 90^\circ$ and $BX$ bisects $\angle KBE$, we have harmonic range $(KXEI)$. Similarly, since $\angle YCI = 90^\circ$ and $CI$ bisects $\angle KCF$, we have harmonic range $(KYFI)$. Finally, $(KEIF)$ is harmonic since $\angle KAI = 90^\circ$ and $AI$ bisects $\angle EAF$. We will now show that $(KXIY)$ is harmonic with the following general claim. Claim. Suppose $K$, $X$, $E$, $I$, $F$, and $Y$ lie on a line such that $(KXEI)$, $(KYFI)$, and $(KEIF)$ are harmonic. Then $(KXIY)$ is harmonic. Proof. Consider a projective transformation that sends $I$ to infinity. Then under this transformation, \[ KX = XE, \; KY = YF, \; KE = KF. \]We need to show that $KX = KY$. But this follows from $KE = 2KX$ and $KF = 2KY$. Since $KE$ and $KF$ are equal, so must be $KX$ and $KY$. We return to the problem. Since $(KXIY)$ is harmonic and $\angle KAI = 90^\circ$, it follows that $AI$ bisects $\angle XAY$. Since $AI$ also bisects $\angle BAC$, we must have $\angle BAX = \angle CAY$, as desired.

Nice problem. Any line passing through $K$ works for $KI$.

let $I_a$ the $A$-excenter then $-1=I_a(B,C;I,K)=(X,Y;I,K)$ but $\angle IAK$ is right angled thus $AI$ is bisector of $\angle XAY$ therefore the result follows RH HAS

Let $I_A$ be the $A$-excenter and notice $$-1=(K,\overline{AI}\cap\overline{BC};B,C)\stackrel{I_A}=(K,I;X,Y)$$by the Right Angles and Bisectors Lemma. The same lemma shows that $\overline{AI}$ bisects $\angle XAY.$ $\square$

I spent wayyy too much time on this lol, same as the above solution. Let $I_A$ be the $A$-Excenter of $\triangle ABC$. Notice that $-1=(K,AI_A \cap BC;B,C) \stackrel{I_A}{=} (K,I;X,Y)$ which combined with the Right angles and bisectors lemma finishes the problem $\blacksquare$ 1 day for such a short solution -__-

Rephrase the problem by considering $\Delta ABC$ as the orthic triangle of $\Delta I_AI_BI_C$ (where $I_A$ is the $A$-excentre, etc.): Quote: Let $H$ be the orthocenter of $\Delta ABC$, $D=AH \cap BC$, $K$ be the $A$-Ex point. $KH$ intersects $AB,AC$ at $X,Y$ respectively. Show that $AD$ bisects $\angle XDY$. Now note that $-1=(B,C;D,K) \stackrel{A}{=} (X,Y;H,K) \stackrel{D}{=} (X,YD \cap AB; A,B)$, and since $AD \cap BC$, the result follows.