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)?

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.

• That's perfect but what about obtuse triangles? @richrow Aug 1 '20 at 3:10
• Actually, obtuse triangle has two acute angles, so we can apply this argument for $AOH$ and $BOH$ and obtain $AH=BH$ if $\angle C>90^{\circ}$. I will edit my answer later. Aug 1 '20 at 5:08