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In geometry, Euler's theorem, named after Leonhard Euler, states that the distance d between the circumcentre and incentre of a triangle can be expressed as

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

where R and r denote the circumradius and inradius respectively (the radii of the above two circles).

From the theorem follows the Euler inequality:

$$R \ge 2r.$$

Proof
A figure for following the proof (which also contains the proof here). Made in GeoGebra software.

Let 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 mid-point 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, then 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

angle BIL = angle A / 2 + angle ABC / 2,

angle IBL = angle ABC / 2 + angle CBL = angle ABC / 2 + angle A / 2,

therefore angle BIL = angle 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).

Bicentric quadrilateral#Fuss' theorem and Carlitz' identity for the relation among the same three variables in bicentric quadrilaterals

Euler's theorem on MathWorld

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