Edmonds, Allan L.; Hajja, Mowaffaq; Martini, Horst (2008), "Orthocentric simplices and biregularity", Results in Mathematics, 52 (1-2): 41–50, doi:10.1007/s00025-008-0294-4, MR 2430410, "It is well known that the incenter of a Euclidean triangle lies on its Euler line connecting the centroid and the circumcenter if and only if the triangle is isosceles."

See here: https://artofproblemsolving.com/community/q1h1621342p10147701

Solution 1 (Barycentric Coordinates). Use the sidelength form of $H$. The determinant of $G, H, I$ simplifies as $$(a - b) (a - c) (b - c) (a + b + c)^2 = 0 \rightarrow a=c \text{ OR } b=c \text{ OR } a=b,$$which finishes. Solution 2 (Complex Numbers). Set $(ABC)$ as the unit circle and $A=a^2, B=b^2, C=c^2$. By Euler Line it suffices to show $H,I,O$ collinear which means we must show $\frac{I}{H} \in \mathbb{R}$. We have $H=a^2+b^2+c^2, I=-(ab+bc+ac)$ and expanding the conjugate and setting it equal gives the desired result upon some algebra.