Let $ABC$ be a scalene acute triangle, where $AL$ $(L \in BC)$ is the internal bisector of $\angle BAC$ and $M$ is the midpoint of $BC$. Let the internal bisectors of $\angle AMB$ and $\angle CMA$ intersect $AB$ and $AC$ in $P$ and $Q$, respectively. Prove that the circumcircle of $APQ$ is tangent to $BC$ if and only if $L$ belongs to it.
Note that $\frac{AP}{PB} = \frac{AM}{MB} = \frac{AM}{MC} = \frac{AQ}{QC}$ so $PQ \parallel BC$ so $APQ$ and $ABC$ are tangent. It's well-known that if $APQ$ is tangent to $BC$ at $L'$ then $AL'$ passes through midpoint of arc $BC$ so $L$ and $L'$ are the same. Now assume that $L$ belongs to $APQ$ and $APQ$ intersects $BC$ at $K$ for the second time. Let $T$ be the midpoint of arc $BC$. Let $TK$ meet $ABC$ again at $R$. Note that $ALKR$ would be cyclic so $APQ$ meets $ABC$ again at $R$ which gives contradiction since they are tangent.