Euler's formula OI² = R² − 2Rr

Step 1: $MB=MI$. In triangle $ABI$ the exterior angle at $I$ gives $\angle BIM=\angle IAB+\angle IBA=\frac A2+\frac B2$. On the other hand $\angle IBM=\angle IBC+\angle CBM=\frac B2+\angle CAM=\frac B2+\frac A2$. So triangle $MBI$ is isosceles with $MB=MI$.

Step 2: lengths. In the circumcircle the chord $MB$ subtends the inscribed angle $\frac A2$, so $MB=2R\sin\frac A2$. In the right triangle formed by $I$, the foot of the perpendicular from $I$ to $AB$, and $A$, we have $AI=\dfrac r{\sin\frac A2}$.

Step 3: power of the point. The point $I$ is inside the circumcircle and the chord through $I$ is $AM$, so
$$R^2-OI^2=IA\cdot IM=\frac r{\sin\frac A2}\cdot2R\sin\frac A2=2Rr.$$
Hence $OI^2=R^2-2Rr$.

Consequence. $OI^2\ge0$ gives $R\ge2r$, with equality iff $O=I$, i.e. iff the triangle is equilateral.

Check. For the $3$-$4$-$5$ triangle, $R=2.5$ and $r=1$, so $OI^2=1.25$; coordinates give the same ✓.

Lesson. Compute a power of a point in two ways, and use the incentre–excentre lemma ($MB=MI$) to link the incentre to the circumcircle.