, 2012

Let $I$ be the incenter of $ABC$, $H_a$ the orthocenter of $I_aBC$, and $O_a$ the midpoint of the segment $II_a$.

We have
$$BO_a = IO_a = I_aO_a = CO_a,$$
so $O_a$ is the circumcenter of triangle $I_aBC$.

Moreover, $IB \parallel CH_a$ (both are perpendicular to $BI_a$) and, similarly, $IC \parallel BH_a$, hence $BICH_a$ is a parallelogram.

Let $P$ denote the projection of $I_a$ onto $BC$ and $T$ the midpoint of segment $AI_a$. We have $H_aP = r$, hence
$$\frac{IA}{AI_a} = \frac{r}{r_a} = \frac{H_aP}{I_aP}.$$
Therefore
$$\frac{H_a I_a}{I_a P} = \frac{I I_a}{I_a A} = \frac{2 I_a O_a}{2 I_a T} = \frac{I_a O_a}{I_a T}. \quad (1)$$
The relation (1) proves that $O_a H_a$ is parallel to $TP$. But $TP \parallel AM$, hence $AM \parallel O_a H_a$, and we are done.

We will use the notation in the previous solution. The quadrilateral $BICI_a$ is cyclic, inscribed in a circle of diameter $II_a$, which implies $O_a$ is the midpoint of segment $II_a$. Also, we have that $IBH_aC$ is a parallelogram.

We want to prove that
$$\frac{I_a O_a}{I_a A} = \frac{I_a H_a}{I_a M}. \qquad (1)$$
We have
$$\frac{IA}{II_a} = \frac{K[BIA]}{K[BII_a]} = \frac{c \sin \frac{B}{2}}{BI_a}.$$
Hence
$$\frac{AO_a}{O_a I_a} = \frac{IA + \frac{1}{2}II_a}{\frac{1}{2}II_a} = \frac{c \sin \frac{B}{2} + \frac{1}{2}BI_a}{\frac{1}{2}BI_a} = 1 + \frac{c \sin \frac{B}{2}}{\frac{1}{2}BI_a}, \quad (2)$$
and
$$\frac{TH_a}{H_a I_a} = \frac{TC \sin \frac{B}{2}}{CI_a \sin \frac{A}{2}}.$$
The relation (3) implies
$$\frac{MH_a}{H_a I_a} = 1 + \frac{2TC \sin \frac{B}{2}}{CI_a \sin \frac{A}{2}}. \qquad (4)$$
We have
$$\frac{c \sin \frac{B}{2}}{BI_a} = \frac{TC \sin \frac{B}{2}}{CI_a \sin \frac{A}{2}} \Leftrightarrow \frac{c \sin \frac{A}{2}}{TC} = \frac{BI_a}{CI_a} = \frac{\sin\left(90^\circ - \frac{C}{2}\right)}{\sin\left(90^\circ - \frac{B}{2}\right)} \Leftrightarrow$$
$$\frac{c \sin \frac{A}{2}}{s-b} = \frac{\cos \frac{C}{2}}{\cos \frac{B}{2}} \Leftrightarrow \frac{c \sqrt{\frac{(s-b)(s-c)}{bc}}}{s-b} = \frac{\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{s(s-b)}{ac}}} \Leftrightarrow$$
$$\frac{\sqrt{c}}{\sqrt{s-b}} = \frac{\sqrt{c}}{\sqrt{s-b}},$$
a relation which is obviously true. It follows that
$$\frac{AO_a}{O_a I_a} = \frac{MH_a}{H_a I_a},$$
and therefore $AM \parallel O_aH_a$.