where R and r denote the circumradius and inradius respectively (the radii of the above two circles).
From the theorem follows the Euler inequality:
<math>R ge 2r. </math>
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
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. d<sup>2</sup> = R(R − 2r).