Prove that the midpoint of $AH$, $K$ and the midpoint of $BC$ are collinear.

$$H$$ 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.

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