$H$ and $K$ are respectively the orthocenter of and a point inside $\triangle ABC$ $(H \not\equiv K)$. $M$, $N$ and $P$ are respectively the orthocenters of $\triangle HKA$, $\triangle HKB$ and $\triangle HKC$. $D$, $E$ and $F$ are respectively the midpoints of $AM$, $BN$ and $CP$. Prove that $D$, $E$ and $F$ are collinear.
It can be observed straight away that $AM \parallel BN \parallel CP$ $($since they are all perpendicular to $HK)$.
It can also be deduced that $KM \parallel BC$, $KN \parallel CA$ and $KP \parallel AB$.
Also, if I let $A'$, $B'$ and $C'$ respectively be the midpoint of $AH$, $BH$ and $CH$ then $(A'B'C')$ is the Euler circle of $\triangle ABC$.
And I'm thinking of a transformation that could do $(A'B'C') \longleftrightarrow \overline{D, E, F}$ potentially.