Geometry problem involving orthocentre and midpoint of sides.

Question

Let $AA_1, BB_1, CC_1$ be the altitudes of $\Delta ABC$ and let $AB \neq BC$. If $M$ is midpoint of $BC$, $H$ the orthocentre of $\Delta ABC$ and $D$ the intersection of $BC$ and $B_1C_1$, prove that $DH \perp AM$

I was trying to come up with a pure geometric proof but haven't got far. These are the following approaches I have come up with:-

• The given conclusion is equivalent to the fact that $H$ is the orthocentre of $\Delta AMD$ which is equivalent to $AD \perp HM$
• The only viable way I could think of to use the fact that $M$ is the midpoint of $BC$ is to consider the nine-point circle of $\Delta ABC$. This coupled with the fact that $D$ lies on the common chord (radical axis) of the nine point circle and the diametric circle of $AH$ might be helpful.
• I could also see a lot of perpendicular lines in the figure. This might point to an application of Simson's Theorem.

I would like some hints on how to approach the problem. If possible, please try to provide hints that build upon the above approaches.

Answer 1

Let $X$ be the intersection of $AM$ and $DH$.

The circumcircle $\Gamma_A$ of $B_1 C_1 H$ goes through $A$ since $\widehat{HC_1 A}=\widehat{HB_1 A}$. Its center is the midpoint of $AH$. In a similar way, the circumcircle $\Gamma_{A'}$ of $BCH$ goes through $A'$, i.e. the symmetric of $A$ with respect to $M$. Since $BCB_1 C_1$ is a cyclic quadrilateral, $D$ belongs to the radical axis of $\Gamma_A$ and $\Gamma_{A'}$, that is perpendicular to the line joining the centre of $\Gamma_A$ with the centre of $\Gamma_{A'}$. We know where these centres lie (at the midpoints of $HA$ and $HA'$), hence we know that $DH\perp AM$. In particular, $X=DH\cap AM$ lies both on $\Gamma_A$ and on $\Gamma_{A'}$.

Answer 2

This answer uses reciprocation with respect to a circle and a degenerated version of Pascal's theorem.

LEMMA: Suppose points $A,B,C,D$ lie on a circle in this order. Let $AC,BD$ intersect at $G$, $AB,CD$ at $F$ and $AD,BC$ at $E$. Then line $EF$ is the polar of $G$ with respect to this circle.

PROOF: Using duality, it is enough to show that $G$ lies on polars of $E$ and $F$. I will show this for $F$ only.

Let tangents to the circle at $A,B$ intersect at $H$ and tangents at $C,D$ at $I$. It's easy to see that line $AB$ is the polar of $H$ and $CD$ is the polar of $I$. Since these polars intersect at $F$, by duality the polar of $F$ passes through $H,I$. So it's enough to show that $E,G,H,I$ are collinear.

By using Pascal's theorem on (degenerate) hexagon $AACBBD$ gives us that $E,G,H$ are collinear. Hexagon $CCADDB$ gives that $I,G,E$ are collinear. This is what we wanted, and this ends proof of the lemma.

Applying this lemma to the circle passing through $BCB_1C_1$ we get that the polar line of $H$ is $AD$, hence $AD$ is perpendicular to the line through $H$ and the center of considered circle (by definition of polar line). But the center is $M$, so $HM\perp AD$, which ends the proof as you have remarked.