$H$ is the orthocentre of $\triangle ABC$. The intersection of the bisectors of $\widehat{ABH}$ and $\widehat{ACH}$ is $K$. Prove that the midpoint of $AH$, $K$ and the midpoint of $BC$ are collinear.
This problem is adapted from a recent competition(, probably the easiest one out there). There might be "in-corrections" that need to be fixed.
Let $BH \cap AC = D$ and $CH \cap AB = E$, the midpoint of $AH$ and $BC$ be respectively $F$ and $G$.
From right-angled triangles $ADH$ and $AEH$, we have that $DF = EF = \dfrac{AH}{2}$.
In addition, from right-angled triangles $BDC$ and $BEC$, we have that $GD = GD = \dfrac{BC}{2}$.
$\implies FG$ is the perpendicular bisector of $DE$.
Furthermore, it is evident that $\widehat{ABD} = \widehat{ACE} ( = 90^\circ - \hat{A}) \implies \dfrac{\widehat{ABD}}{2} = \dfrac{\widehat{ACE}}{2}$
$\implies \left\{ \begin{align} \widehat{DBK} = \widehat{DCK}\\ \widehat{EBK} = \widehat{ECK} \end{align} \right. \implies BCDK$ and $CBEK$ are cyclic quadrilaterals.
$\implies BCDKE$ is a cyclic pentagon. In $(B, C, D, K, E)$, it can be seen that $\widehat{KBD} = \widehat{KCE}$.
$\implies KD = KE \implies K$ lies on the perpendicular bisector of $DE$ or $K \in FG$.
Let the midpoint of $BC$ be $M$. By straightforward angle chase, we obtain $\angle BKC = 90^\circ$. So $K$ is on the circle with diameter $BC$, and so $M$ is the centre of this circle. Note that $\angle EGK = 2 \angle ECK = 2 \angle KBD = \angle KMD$, so $KM$ is the perpendicular bisector of $DE$.
Also note that $ADHE$ is cyclic (due to right angles) and therefore has circumcenter $F$ (midpoint of diameter). So $FE = FD$, and so $F$ lies on perpendicular bisector of $DE$. So $F, K, M$ are collinear.
