The triangle does not have to be isosceles. See Triangles with common Circumcircle, Incircle [Inversion?] for a proof of Euler's formula by inversion.

$\triangle ABC,\ w=C(O,R),\ \{A,A'\}\subset AI\cap w.$ $IA=\frac{r}{\sin \frac A2},\ IA'=BA'=CA'=2R\sin \frac A2\Longrightarrow$ $R^2-OI^2=-p_w (I)=IA\cdot IA'=2Rr\Longrightarrow OI^2=R^2-2Rr.$ Remark. $p_w (I)$ is the power of point $I$ w.r.t. the circumcircle $w$.

Check here for proof: https://www.cut-the-knot.org/triangle/EulerIO.shtml

ZETA_in_olympiad wrote: Euler's formula holds for all triangle, so I'll just prove Euler's formula in the classical way: $\textbf{LEMMA: For all real } x, \textbf{ we have, } e^{ix}=\cos x+i\sin x.$ Proof. Consider the function $$f(\theta)= e^{-i\theta} (\cos \theta+ i\sin \theta).$$Differentiating $f$ yields $$e^{-i\theta} (i\cos \theta-\sin \theta)- ie^{-i\theta} (\cos \theta +i\sin \theta)=0.$$Therefore, $f$ is constant and our claim follows. $\blacksquare$ Ummm this isn't the euler's formula the problem is talking about, you might want to check the problem statement again.

BVKRB- wrote: Ummm this isn't the euler's formula the problem is talking about, you might want to check the problem statement again. It's the special case of Euler's formula for isosceles triangles. Please see #2.

ZETA_in_olympiad wrote: BVKRB- wrote: Ummm this isn't the euler's formula the problem is talking about, you might want to check the problem statement again. It's the special case of Euler's formula for isosceles triangles. Please see #2. The euler's formula #2 was talking about is that for any triangle $ABC$, the distance from incenter to circumcenter is $R(R-2r)$. (Or is this a troll?)

Quidditch wrote: ZETA_in_olympiad wrote: BVKRB- wrote: Ummm this isn't the euler's formula the problem is talking about, you might want to check the problem statement again. It's the special case of Euler's formula for isosceles triangles. Please see #2. The euler's formula #2 was talking about is that for any triangle $ABC$, the distance from incenter to circumcenter is $R(R-2r)$. (Or is this a troll?) I apologize I quoted the wrong Euler's formula. No troll though.