$H$ 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.


1 Answer 1


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.

  • 1
    $\begingroup$ +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, 2020 at 18:03
  • 1
    $\begingroup$ Thanks for posting the figure! The solution is much easier to follow now. $\endgroup$
    – timon92
    Feb 5, 2020 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.