Let $BX\cap CY=T,XY\cap BC=D$. $\left.\begin{array}{lll} \text{paskal's theorem for}\ YYXBBC\Longrightarrow T,N,D\ \text{are collinear}\\ \\ \text{paskal's theorem for}\ XXYCCB\Longrightarrow T,M,D\ \text{are collinear} \end{array}\right\}\Longrightarrow T,N,M,D\ \text{are collinear}\Longrightarrow D\in II_a$ Hence: $\left.\begin{array}{ccc} B,Y,C,X\in (O)\Longrightarrow BD.DC=DY.DX\\ \\ C,B,I,I_a\in (\text{circle with diameter} II_a)\Longrightarrow BD.DC=DI.DI_a \end{array}\right\}\Longrightarrow DI.DI_a=DY.DX\Longrightarrow X,Y,I,I_a\ \text{lie on a circle}$ Q.E.D

Attachments:

Let $AD\cap \odot ABC=\{F\}$. $XX,CC,AF$ are concurrent $\implies$ $AXFC$ is harmonic,analogously so is $ABFY$.Consider an inversion $\Psi$ with center $A$ radius $\sqrt{AB\cdot AC}$ composed with reflection over $A$-angle bisctor.Let $\Psi(Z)=Z^*$. Obviously $\Psi(B)=C$ and $\Psi(I)=I_A$.Inversion preserves harmonic ratios, as $(A,F;X,C)=-1$ $\implies$ $(\infty,F^*;X^*,B)=-1$ $\implies$ $F^*$ is a midpoint of $BX^*$ but $F^*\equiv D$ $\implies$ $BD=DX^*$ ,analogously $CD=DY^*$. $IBI_AC$ is cyclic $\implies$ $ID\cdot DI_A=BD\cdot CD=DX^*\cdot DY^*$ $\implies$ $IX^*I_AY^*$ is cyclic hence we're done.

bgn wrote: Given $\triangle ABC$ inscribed in $(O)$ an let $I$ and $I_a$ be it's incenter and $A$-excenter ,respectively. Tangent lines to $(O)$ at $C,B$ intersect the angle bisector of $A$ at $M,N$ ,respectively. Second tangent lines through $M,N$ intersect $(O)$ at $X,Y$. Prove that $XYII_a$ is cyclic. Let $D$ be the foot of $A$-angle bisector and $L$ be the mid-point of arc $BC$. Then we claim that $\overline{XY}$ passes through $D$. If true, this tells us that $$DX \cdot DY=DA \cdot DL=DI \cdot DI_a$$since $(AD, II_a)=-1$ and $L$ is the midpoint of $II_a$; thus proving that $XYII_a$ is cyclic. Apply inversion at $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in line $\overline{AI}$. Let $\omega_B, \omega_C$ be the circles through $B', C'$ respectively, tangent to line $B'C'$ and passing through $A$. Then $M'$ lies at the intersection of $\overline{AI}$ and $\omega_B$ while $N'$ lies at the intersection of $\omega_C$ and $\overline{AI}$. The other tangents map to the other circles passing through $A$ and touching line $\overline{B'C'}$. Thus, $L'X'^2=L'M' \cdot L'A=L'B'^2 \implies L'X'=L'B'$ and similarly $L'C'=L'Y'$. Thus, $$L'A' \cdot L'D'=L'B' \cdot L'C'=L'X' \cdot L'Y'$$proving $AX'D'Y'$ cyclic; thus $D$ lies on $\overline{XY}$.

Pole-Polar kills this problem by considering the common point of BY and CX along within proving the famous claim(previous Aopsers did that) by Brocard.

Let $XY\cap BC=D$, but consider complete quadrilateral $XYBCD\cup \{Z=BX\cap CY\}$, $M, N$ lies on the line $DZ$, so $D$ lies on $AI_a$. Then with $(BI_aCI), (BCXY)$ we conclude that $XYII_a$ is cyclic by converse of Radical Axis Theorem.

Let $S=XY \cap BC $ and $K=AI \cap (ABC)$ and $R=AA \cap KK$ we should prove that $S \in AD$ if it does then $SX.SY=SB.SC=SI.SI_a$ (the last equality comes from $BCII_a$ is cyclic) claim: polar $M=XC$ polar $N=YB$ then the pole of $MN$ is $XC \cap YB$ and also pole $AK=R \implies R=XC \cap YB$ the new problem : let $ABKC$ a quadrilateral and Let $R=AA \cap KK$ $X=RB \cap (ABKC)$ $Y=RC \cap (ABCD)$ prove that $BC,AK,XY$ are concurrent take a projective transformation that fixes $ABKC$ and send $AK \cap BC$ to the center of $(ABKC)$ $O$ then $ABKC$ is a rectangle and $AA \cap KK=P_\infty$ easily get $X,Y,O$ are collinear