My solution: Let $ T $ be the midpoint of $ AI $ . Since $ T $ is the center of $ \odot (AEZ) $ , so $ \angle ZTE=\angle BOC=2\angle BAC \Longrightarrow OBCM \sim TZEI $ . Since $ K $ is the center of spiral similarity that maps $ BC \mapsto ZE $ , so $ K $ is the center of spiral similarity that maps figure $ OBCM \mapsto $ figure $ TZEI $ . From $ \angle KNO=\angle KMO=\angle KIT=\angle KNA \Longrightarrow O \in AN $ , so combine with $ \angle AKI=90^{\circ} $ we get $ KI \cap AN $ is the antipode of $ A $ in $ \odot (O) \equiv (c) $ . Q.E.D

It's just a trivial angle chasing: Solution: let $ KI $ intersect $\odot (\triangle ABC)$ at $S$ and $AS$ intersect $\odot (\triangle AIK)$ at $ N'$ then we have to show that quadrilateral $ KOMN'$ is cyclic note that $\angle AKI=90$ so $ A, O, S, N'$ are collinear also we have $\angle AMS=\angle AN'I=90$ so $SMIN'$ is cyclic using this fact we obtain that $\angle MN'A=\angle KIA=\angle KN'A$ so $\angle KN'O=\angle MN'O$ note that in the quadrilateral $KOMN$ $ON'$ is bisector of $\angle MNK$ and $OK=OM$ using sin lemma in $\triangle OMN', \triangle ON'K$ we get $\text{sin} OKN'=\text{sin} OMN'$ so $ OKN+OMN=180$ and we are DONE.

Greece TST 2015 P3 wrote: Let $ABC$ be an acute triangle with $\displaystyle{AB<AC<BC}$ inscribed in circle $ \displaystyle{c(O,R)}$.The excircle $\displaystyle{(c_A)}$ has center $\displaystyle{I}$ and touches the sides $\displaystyle{BC,AC,AB}$ of the triangle $ABC$ at $\displaystyle{D,E,Z} $ respectively.$ \displaystyle{AI}$ cuts $\displaystyle{(c)}$ at point $M$ and the circumcircle $\displaystyle{(c_1)}$ of triangle $\displaystyle{AZE}$ cuts $\displaystyle{(c)}$ at $K$.The circumcircle $\displaystyle{(c_2)}$ of the triangle $\displaystyle{OKM}$ cuts $\displaystyle{(c_1)} $ at point $N$.Prove that the point of intersection of the lines $AN,KI$ lies on $ \displaystyle{(c)}$. Solution: Rename $I$ as $I_A$ and instead let $I$ be incenter. Let $A'$ be the $A-$antipode in $\odot (ABC)$. Let $I_AA'$ $\cap$ $\odot (ABC)$ $=$ $K'$ $\implies$ $\angle AMA'$ $=$ $90^{\circ}$ $=$ $\angle AK'I_A$ $\implies$ $K'$ $\in$ $\odot (AZE)$. Let $AA' $ $\cap$ $\odot (AZE)$ $=$ $N'$ $\implies$ $MA'N'I_A$ is cyclic $$\angle A'MN'=\angle KI_AN'=\angle N'AK=\angle A'MK \implies \angle KON'=2\angle KMA'=\angle KMN' \implies N \equiv N'$$Hence, $KI_A \cap AN=A' \in \odot (ABC)$