# Prove that the midpoints of $A$, $B$, $C$ and the orthocenters of $\triangle HKA$, $\triangle HKB$, $\triangle HKC$ are collinear.

$$H$$ 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.

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 $$ABC$$. 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.
• +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. – Blue Feb 5 '20 at 18:03