Euler's theorem in geometry

Remarks

Let the line $AI$ meet the circumcircle again at $M$, the midpoint of the arc $BC$ not containing $A$. An angle chase shows $MI = MB$, and the law of sines in the triangle $ABM$ gives $MB = 2R \sin \frac{A}{2}$. Also $AI = r / \sin \frac{A}{2}$, so $AI \cdot IM = 2Rr$. The power of the point $I$ with respect to the circumcircle is $R^2 - d^2 = AI \cdot IM$, and the formula follows.

The inequality $R \ge 2r$ is Euler's inequality. Equality $d = 0$ means that the circumcentre and the incentre coincide, which happens only for an equilateral triangle.

Example: for the right triangle with sides $3$, $4$ and $5$, the circumradius is $R = \frac{5}{2}$ and the inradius is $r = 1$. The formula gives $d^2 = \frac{5}{2} \cdot \frac{1}{2} = \frac{5}{4}$. With $C = (0, 0)$, $A = (4, 0)$, $B = (0, 3)$ one has $O = (2, \frac{3}{2})$ and $I = (1, 1)$, and indeed $OI^2 = 1 + \frac{1}{4} = \frac{5}{4}$.

Going deeper

For a student

The theorem is a basic tool for inequalities in triangle geometry, through $R \ge 2r$. It also enters the proof of Feuerbach's theorem via the distance between the incentre and the nine-point centre. Its converse is a closure statement. For two circles with $d^2 = R(R - 2r)$, every point of the outer circle is a vertex of a triangle inscribed in one and circumscribed about the other. This is the case of triangles in Poncelet's closure theorem.