Thai Mathematical Olympiad

Consider the following well known Lemma:

Lemma. Let $ABC$ be a triangle with incircle $I$. If $M$ is the midpoint of the arc $BC$ of the circumcircle not containing $A$, then $MB=MC=MI$.

Proof of Lemma. Let the circumcircle of $DEF$ intersect $\Omega$ again at $X$. Let $M$ and $N$ be the midpoints of the arcs $BC$ (not containing $A$) and $AC$ (not containing $B$), respectively. Let $P$ be the intersection of $\Omega$ and the line through $C$ parallel to $MN$ (if $P$ is the same as $C$, i.e., $AC=CB$, the results below still hold.) $\square$

By above Lemma and since $MNCP$ is an isosceles trapezoid, we get $MP=NC=NE$ and $NP=MC=MF$. Since $E$ and $F$ lie on $DN$ and $DM$ respectively, $\angle NXM=\angle NDM=\angle EDF=\angle EXF$. Therefore, $\angle NXE=\angle MXF$ and since $\angle XNE=\angle XND=\angle XMD=\angle XMF$, we have that $△NXE\sim △MXF$. Then, $\frac{NX}{NE}=\frac{MX}{MF}$, and since $NE=MP$ and $MF=NP$, we get $$NX\cdot NP=MX\cdot MP.$$ Therefore, $\frac{[NPX]}{[MPX]}=\frac{NX\cdot NP}{MX\cdot MP}=1$. This implies that the line $XP$ bisects the segment $MN$. Therefore, $X$ must lie on the intersection of $\Omega$ and the line joining $P$ and the midpoint of $MN$. Since $M,N,P$ are fixed independent of $D$, therefore, $X$ is the only loci as $D$ varies on the arc $AB$.