Prove excircles are on circumcircle with incircle 
Let ABC be a triangle. The $A-$ and $B-$ angle bisectors
intersect opposite sides at $K, L$, respectively, and intersect each other at $I$. Line
KL intersects the circumcircle $\omega$ of $\triangle ABC$ at $X$ and $Y$. Prove that the circumcircle
of $\triangle IXY$ passes through the $A-$ and $B-$ excenters of $\triangle ABC$.

I've attempted the problem by considering the radical axis of the circumcircle of $\triangle ABC$ and the $B-$excircle, as well as with the $A-$excircle to find any cyclic quadrilaterals, but no luck there. I have also tried to find any similar triangles, but I can't finding anything useful and I'm quite stuck. Any hints on how to proceed?
 A: Notations. Let in $\triangle ABC$, $I$ be the incenter, $I_A,I_B,I_C$ be respective excenters. Let $\{AI\cap BC,BI\cap AC,CI\cap AB\}\equiv\{J_A,J_B,J_C\}$. Let $J_AJ_B\cap \odot(ABC)=\{L_C, L'_C\}$, $J_AJ_C\cap \odot(ABC)=\{L_B, L'_B\}$ and $J_CJ_B\cap \odot(ABC)=\{L_A, L'_A\}$.
Proof. Thus, we need to show that $IL_CL'_CI_AI_B$ is a cyclic pentagon. Clearly, its enough to show that $IL_CL'_CI_A$ is cyclic quadrilateral as the other part follows by symmetry. From incenter-excenter lemma, we know that $IBI_AC$ is cyclic and thus,
$$J_AI\cdot J_AI_A\overset{\mathcal P_{\odot(IBC)}(J_A)}{=} J_AB\cdot J_AC\overset{\mathcal P_{\odot(ABC)}(J_A)}{=} J_AL_C\cdot J_AL'_C$$and thus, by converse of power of point theorem, we get $IL_CL'_CI_A$ cyclic which completes the proof. $\square$
A: I'm still working on the solution, however, I've solved the problem. Proving your statement is equivalent to proving that : The circumradius of $\triangle IXY$ is twice the circumradius of $\triangle ABC$. Alternatively, we can also prove that the circumcircle of $\triangle IA'B'$ passes through $X$ and $Y$.
What I noticed that $\omega$ is nine-point circle of $\triangle IA'B'$ and $\triangle ABC$ is pedal triangle of $\triangle IA'B'$.

You were going on somewhat a right track. It has something to do with the radical axis.
After it is known that $\triangle ABC$ is orthic (pedal) triangle of $\triangle IA'B'$, we can say that $KL$ is the orthic axis of $\triangle IA'B'$ which coincides with radical axis of its circumcircle and $\omega$.
Also, as $\displaystyle\angle A'IB'>\frac{π}{2}$, the two circles do intersect. So, the two points of intersection are definitely $X$ and $Y$.
