# Euler's theorem in geometry

Euler's theorem:
${\displaystyle d=|IO|={\sqrt {R(R-2r)}}}$

In geometry, Euler's theorem states that the distance d between the circumcentre and incentre of a triangle is given by[1][2][3][4]

${\displaystyle d^{2}=R(R-2r)\,}$

or equivalently

${\displaystyle {\frac {1}{R-d}}+{\frac {1}{R+d}}={\frac {1}{r}},}$

where R and r denote the circumradius and inradius respectively (the radii of the circumscribed circle and inscribed circle respectively). The theorem is named for Leonhard Euler, who published it in 1767.[5] However, the same result was published earlier by William Chapple in 1746.[6]

From the theorem follows the Euler inequality:[2][3]

${\displaystyle R\geq 2r,}$

which holds with equality only in the equilateral case.[7]:p. 198

## Proof

Proof of Euler's theorem in geometry

Letting O be the circumcentre of triangle ABC, and I be its incentre, the extension of AI intersects the circumcircle at L. Then L is the midpoint of arc BC. Join LO and extend it so that it intersects the circumcircle at M. From I construct a perpendicular to AB, and let D be its foot, so ID = r. It is not difficult to prove that triangle ADI is similar to triangle MBL, so ID / BL = AI / ML, i.e. ID × ML = AI × BL. Therefore 2Rr = AI × BL. Join BI. Because

BIL = ∠ A / 2 + ∠ ABC / 2,
IBL = ∠ ABC / 2 + ∠ CBL = ∠ ABC / 2 + ∠ A / 2,

we have ∠ BIL = ∠ IBL, so BL = IL, and AI × IL = 2Rr. Extend OI so that it intersects the circumcircle at P and Q; then PI × QI = AI × IL = 2Rr, so (R + d)(R − d) = 2Rr, i.e. d2 = R(R − 2r).

## Stronger version of the inequality

A stronger version[7]:p. 198 is

${\displaystyle {\frac {R}{r}}\geq {\frac {abc+a^{3}+b^{3}+c^{3}}{2abc}}\geq {\frac {a}{b}}+{\frac {b}{c}}+{\frac {c}{a}}-1\geq {\frac {2}{3}}\left({\frac {a}{b}}+{\frac {b}{c}}+{\frac {c}{a}}\right)\geq 2,}$

where a, b, c are the sidelengths of the triangle.

## Euler's theorem for the exscribed circle

If ${\displaystyle r_{a}}$ and ${\displaystyle d_{a}}$ denote respectively the radius of the exscribed circle opposite to the vertex ${\displaystyle A}$ and the distance between its centre and the centre of the circumscribed circle, then ${\displaystyle d_{a}^{2}=R(R+2r_{a})}$.