The incenter and Euler line. It seems well known that the incenter of a triangle lies on the the Euler line if and only if the triangle is isosceles (or equilateral, but that is trivial). Searching the internet, I could not find a simple geometric proof of this fact. Can anyone provide such a proof? Also, when the incenter lies on the Euler line, does it do so in a set location? (For example, we know the centroid is a third of the way from the circumcenter to the orthocenter on the Euler line, does the incenter satisfy any nice ratios like that?)
 A: Here's a nice proof by contradiction.
Let the incenter $I$ lie on the Euler line of $ABC$.
It is known that orthocenter $H$ and circumcenter $O$ are isogonal conjugates, i.e. $AI$ is the bisector of angle $HAO$. 
So, (if the point $A$ does not lie on Euler line) $HA/AO=HI/IO$ (angle bisector theorem).
Also $HB/BO=HC/CO=HI/IO=HA/AO$.
And we know that all points $X$, such that $YX/ZX=const$, lie on a circle with center on the line $YZ$ (Appolonius circle)
So, $A, B, C$ and $I$ lies on the same circle, and that cannot be true.
We have assumed that all of points $A, B, C$ don't lie on the Euler line, so, one of them lies on Euler line and that means $ABC$ is isosceles.
A: One approach could be to use trilinear coordinates and show that the incentre at $1:1:1$ is usually not collinear with for example the circumcentre at $\cos A :\cos B :\cos C$ and the orthocentre at $\sec A :\sec B :\sec C$ by looking at the determinate 
$$\begin{vmatrix}1&1&1\\ \cos A &\cos B &\cos C\\ \sec A &\sec B &\sec C\end{vmatrix}$$
which is non-zero unless at least two of $A$, $B$ and $C$ are equal.
A: The Incenter of the ABC triangle of sides a, b and c is at a distance d from the Euler line given by the formula:
$d=\frac{1}{2}\frac{|(a-b)(a-c)(b-c)|}{\sqrt{(abc)^2-(-a^2+b^2+c^2)(a^2-b^2+c^2)((a^2+b^2-c^2)}}$
If the distance is equal to zero the Incenter is on the Euler line. The formula results in zero in the case of the isosceles triangle (a-b = 0 or a-c = 0 or b-c = 0) and an equilateral triangle (a = b = c).
A: I would like to present the result of W. Massaro in a slightly different manner.
It is known (see for example p. 82 of his document](http://www.journal-1.eu/2017/Grozdev-Okumura-Dekov-Euler-Line-pp.81-85.pdf)) that the Euler line of a triangle with sidelengths $a,b,c$ has barycentric equation:
$$(b^2-c^2)(b^2+c^2-a^2)x+(c^2-a^2)(c^2+a^2-b^2)y+(a^2-b^2)(a^2+b^2-c^2)z=0\tag{1}$$
As the barycentric coordinates of the incenter are $(x,y,z)=(a,b,c)$, plugging them into (1) results in an expression which, factorized (using a Computer Algebra System !), becomes:
$$(a-b)(b-c)(c-a)(a+b+c)^2$$
Nothing astonishing that we find back the expression in the numerator of the result of W. Massaro.
