Problem - 2435
Let $O$ be the incenter of $\triangle{ABC}$. Connect $AO$, $BO$, and $CO$ and extends so that they intersect with $\triangle{ABC}$'s circumcircle at $D$, $E$, and $F$, respectively. Let $DE$ intersect $AC$ at $G$, and $DF$ intersects $AB$ at $H$. Show that $G$, $H$ and $O$ are collinear.
It is sufficient to show that $OG\parallel BC$ and $OH\parallel BC$.
First, let's connect $OG$ and $OG$. Because $O$ is the incenter of $\triangle{ABC}$, therefore $\angle{BAD}=\angle{CAD}$ which means are $\stackrel{\frown}{BD}=\stackrel{\frown}{DC}$. Similarly, we have $\stackrel{\frown}{CE}=\stackrel{\frown}{EA}$ and $\stackrel{\frown}{AF}=\stackrel{\frown}{FB}$. Therefore, $$\stackrel{\frown}{BD}+\stackrel{\frown}{AE}+\stackrel{\frown}{AF}=180^\circ\implies \angle{BND}=90^\circ$$
Meanwhile, because $BE$ bisects $\angle{ABC}$, we find $HN=NM$.
Additionally, $\stackrel{\frown}{AF}=\stackrel{\frown}{FB}$ means $FD$ bisects $\angle{BDA}$. This means $BN=NO$. Hence, we find $BMOH$ is a parallelogram which results in $OH\parallel BC$.
By symmetry, $OG\parallel BC$ must hold too.