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$H$ 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.

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enter image description here 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 $K \in FG$.

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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.

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  • $\begingroup$ What is $G$?.......... $\endgroup$ – Aqua Jun 12 at 13:40
  • $\begingroup$ You are pretty ignorant. $\endgroup$ – Aqua Jun 12 at 17:46

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