, 2018

Denote $(I_{a}), (I_{b}), (I_{c})$ as the ex-circles with respect to angles $A, B, C$ of triangle $ABC$. It is easy to check that $\mathscr{P}_{M/(I)} = \mathscr{P}_{M/(I_{a})}$ and $\mathscr{P}_{M/(I_{b})} = \mathscr{P}_{M/(I_{c})}$. Let $O, K$ be the circumcenters of triangles $ABC$ and $DEF$.

Hence, $d_{1}$ is the radical axis of $(I), (I_{a})$ since $d_{1} \perp II_{a}$. Similarly for $d_{2}, d_{3}$, which implies that $D = d_{2} \cap d_{3}$ is the radical center of the three circles $(I), (I_{b}), (I_{c})$. Thus, $MD$ is the radical axis of $(I_{b}), (I_{c})$.

So we have $MD \perp I_{b}I_{c}$, but $I_{b}I_{c} \perp AI$, $EF \perp AI$, then $MD \perp EF$ leads to $MD$ being the altitude in triangle $DEF$.

Similarly for $NE, PF$, so the three lines $MD, NE, PF$ concur at point $T$, which is the orthocenter of $DEF$. The two triangles $ABC$ and $DEF$ share the same nine-point circle $(MNP)$, so $OH$ and $KT$ share the midpoint, which is the circumcenter of triangle $(MNP)$. This implies that $\overrightarrow{HK} = \overrightarrow{TO}$.

On the other hand, it is easy to check that $T$ is also the incenter of triangle $MNP$. Then consider the homothety with center $G$ as the centroid of triangle $ABC$, ratio $-2$, then $\mathscr{H}: M \rightarrow A, N \rightarrow B, P \rightarrow C$ implies that $\mathscr{H}: T \rightarrow I, O \rightarrow H$. Hence, $\overrightarrow{HI} = 2\overrightarrow{TO}$.

Combine with the previous equality, we have $\overrightarrow{HI} = 2\overrightarrow{HK}$, so $K$ is the midpoint of the segment $HI$. $\square$