Ukraine TST 2014 P4 wrote: The $A$-excircle of the triangle $ABC$ touches the side $BC$ at point $K$. The circumcircles of triangles $AKB$ and $AKC$ intersect for the second time with the bisector of angle $A$ at points $X$ and $Y$ respectively. Let $M$ be the midpoint of $BC$. Prove that the circumcenter of triangle $XYM$ lies on $BC$. Key Claim:- $M$ is the Incenter of $\Delta XKY$ Proof:- Notice that $\angle XKM=\angle CAB=\angle XAC=\angle YKM$. Now $$\frac{MD}{MK}=\frac{BM-BD}{CM-CK}=\frac{\left(\frac{a}{2}-\frac{ac}{b+c}\right)}{\left(\frac{b}{2}-\frac{c}{2}\right)}=\frac{a}{b+c}=\frac{1}{c}\cdot\left(\frac{ac}{b+c}\right)=\frac{DB}{AB}=\frac{DX}{XK}$$Hence, $M$ is the Incenter of $\Delta XKY$. Hence Circumcenter of $\Delta XMY\in BC.$ $\blacksquare$

Here’s a pure angle chasing solution. We’re going to prove that $M$ is the incenter of triangle $KXY$. Add incenter $I$. Using the fact that $IM\parallel AK$, we have $$\angle XIM=\angle XAK= \angle XBK =\angle XBM.$$Thus, $B,I,M,X$ are concyclic, implying that $$\angle YXM=\angle IXM=\angle IBM=\frac{\angle ABC}{2}.$$However, $\angle YXK=\angle AXK=\angle ABK=\angle ABC$. So, $XM$ bisects $\angle YXK$. Similarly, $YM$ bisects $\angle XYK$, as desired.