If $R$ is the circumradius and $r$ is the inradius of some triangle $ABC$, with its circumcenter being $O$ and incenter being $I$, then how to prove: $$OI^2=R(R-2r)$$ I have seen many mentions of this theorem, and Euler's inequality is a corollary. Wikipedia has a proof, but its quite tough to follow and long, so does anyone have a nicer proof? :)
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3There are different proofs in https://www.dpmms.cam.ac.uk/~njb65/Euler.pdf – Jaakko Seppälä Oct 14 '14 at 17:24
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@JaakkoSeppälä That is awesome :) – user184352 Oct 14 '14 at 17:33
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2And BTW, the formula is $OI^2=R(R-2r)$ so remember that square term. – Jaakko Seppälä Oct 14 '14 at 17:34
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@JaakkoSeppälä Can you add that as an answer, I would like to accept it? :) – user184352 Oct 14 '14 at 18:12
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1there is one in here,which is originally from maa – user2838619 Oct 14 '14 at 18:17
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1This is a famous result due to Leonhard Euler. It is a matter of taste which proof is the nicest one. You can use for example trigonometry, inversion, or Poncelet's porism. See http://www.dpmms.cam.ac.uk/~njb65/Euler.pdf for details. – Jaakko Seppälä Oct 14 '14 at 18:18
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1In this answer, I derive this relation. – robjohn Oct 24 '14 at 08:40