The incenter and Euler line.

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.

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.

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).