can incentre lie on the Euler line for an obtuse triangle? I know that  incentre lie on the Euler line for equilateral and isosceles triangle but I found a claim
that incentre can lie on the Euler line for obtuse triangle. So, is this claim true?Also does there exist any scalene and acute ( but neither equilateral or isosceles ) triangle for which incentre lies on the Euler line? Finally, if incentre is on Euler line, then is it at a unique location with respect to other centres (orthocentre, circumcentre, centroid)?
 A: Case 1. Acute triangle.
Let $ABC$ be an acute triangle and $I$, $O$, $H$ are its incenter, circumcenter and orthocenter, respectively. Note that points $I$, $O$ and $H$ are inside triangle $ABC$.
We will prove that $I$, $O$ and $H$ are collinear iff $ABC$ is isosceles or equilateral. Indeed, suppose that $O$, $I$ and $H$ are collinear but $\triangle ABC$ is scalene. Recall that rays $AO$ and $AH$ are symmetric with respect to angle bisector of $\angle BAC$. Hence, angle bisectors of angles $OAH$ and $BAC$ coincide, so $AI$ bisects angle $AOH$. Since $I\in OH$ we have
$$
\frac{AO}{AH}=\frac{IO}{IH}
$$
due to angle bisector theorem for $\triangle AOH$.
Similarly, we obtain that
$$
\frac{AO}{AH}=\frac{BO}{BH}=\frac{CO}{CH}=\frac{IO}{IH}.
$$
Finally, note that $AO=BO=CO$, so the last equality implies $AH=BH=CH$. Thus, $O$ and $H$ are distinct circumcenters of triangle $ABC$ which is impossible.
Therefore, in acute triangle $O$, $I$ and $H$ are collinear iff $\triangle ABC$ has equal sides.
Case 2. Obtuse (or right) triangle.
Suppose that in $\triangle ABC$ we have $\angle C\geq 90^{\circ}$. In this case we still can apply the previous argument to triangles $AOH$ an $BOH$ (because rays $AO$ and $AH$ are still symmetric with respect to $AI$; the same for rays $BO$, $BH$ and $BI$). Thus,
$$
\frac{AO}{AH}=\frac{BO}{BH}=\frac{IO}{IH}.
$$
However, it means that $AH=BH$, so $AB=BC$, so triangle $ABC$ is isosceles, as desired.
