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$H$ 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.

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Let $\mathcal H$ be the hyperbola passing through $A, B, C, H, K$. Then $\mathcal H$ is a rectangular hyperbola because $H$ is the orthocenter of $ABC$.

enter image description here

Since the hyperbola $\mathcal H$ is rectangular, it follows that the orthocenters of $HKA$, $HKB$, $HKC$ lie on $\mathcal H$. So, $AM$, $BN$, $CP$ are parallel chords of $\mathcal H$ (as all are perpendicular to $HK$).

Denote by $X$ the (common) point at infinity of the lines $AM$, $BN$, $CP$. Then the quadruples $AMDX$, $BNEX$, $CPFX$ are harmonic. It follows that points $D, E, F$ lie on the polar of $X$ with respect to $\mathcal H$. Hence $D$, $E$, $F$ are collinear.

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    $\begingroup$ +1. Very nice! (I took the liberty of adding a figure.) Note: Once we have that $AM$, $BN$, $CP$ are parallel chords, we "know" that the corresponding midpoints are collinear (and that their common line contains the center of the conic) as a projectively-preserved property of circles. $\endgroup$ – Blue Feb 5 '20 at 18:03
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    $\begingroup$ Thanks for posting the figure! The solution is much easier to follow now. $\endgroup$ – timon92 Feb 5 '20 at 19:30

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