Proof
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 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, 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).
Read more about this topic: Euler's Theorem In Geometry
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