While trying to solve this question using GeoGebra, I realized the following curious thing:
If $I$ is the incenter of $\triangle ABC$, $ID \perp BC$ with $D$ on $BC$, $AD \perp IO$ with $O$ on $AD$, then if $H$ is the orthocenter of $\triangle ABC$, $O$, $I$, $H$ are collinear.
Edit: Sorry people, its wrong and I came up with it by mistake, don't waste your time on it.
On
By using trilinear coordinates (today is the day) we have that a point belongs to the Euler line $OH$ iff its trilinear coordinates $[x:y:z]$ satisfy $$ \sum_{cyc} \sin(2A)\sin(B-C)x = 0. \tag{1}$$ For the incenter $I=[1:1:1]$, the last identity is equivalent to: $$ \left(\sin(A)+\sin(B)+\sin(C)\right)\prod_{cyc}\sin\frac{A-B}{2},\tag{2} $$ so, if we assume that $ABC$ is a scalene triangle, $I$ belongs to the Euler line iff $$ \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} = 0, \tag{3} $$ no way.
My belief is that the statement is false.
Assume it is true, that OIH is a straight line. Then, AOD and IOH are perpendicular, meaning that AD is perpendicular to IH.
Similarly, this implies that BE and CF (where E, F are the foot of the perpendiculars from I to AC, and AB respectively) are perpendicular to IH.
But this is clearly not possible.
Hence, the statement is false.
The following image shows the points A, D, I, H as defined (without the point O as yet). When we add the point O such that $IO \perp AD$, it does not look like $OIH$ is a straight line.