Euler Triangle Formula

Theorem
Let $$d$$ be the distance between the incenter and the circumcenter of some triangle. Then $$d^2=R(R-2 \rho)$$, where $$R$$ is the circumradius and $$\rho$$ is the inradius.

Lemma
If we extend the bisector ,of one angle of the triangle, to one point of the circunference ,let said P.Then the distance of any of the vertices,of the triangle, left are equal to IP, where I is the incenter of some traingle

Proof (Lemma)


in the figure we have that CP si the bisector of $$\angle ACB$$ Now how $$\angle ABP$$ and $$\angle ACP$$ subtend the same arc then $$\angle ACP = \angle ABP = \dfrac{\angle C}{2} $$ Like IB is the bisector of B then $$\angle IBA = \dfrac{\angle B}{2} \Rightarrow \angle IBP = \dfrac{\angle B + \angle C}{2}$$

and in $$\triangle CIB$$ the suplement of$$\angle CIB = \dfrac{\angle C}{2}+\dfrac{\angle B}{2}$$ therefore $$\angle PIB = \angle IBP \Rightarrow IP = BP$$

Proof (Theorem)


$$OI = d$$, $$OG = OJ = R$$.

Therefore $$IJ = R + d$$ and $$GI = R-d$$.

By the Chord theorem we have $$GI \cdot IJ =IP \cdot CI$$.

Now by the lemma we have $$GI \cdot IJ = PB \cdot CI$$.

Now using the Extension of Law of Sines in $$\triangle CPB$$ we have $$\dfrac{PB}{\sin(\angle PCB)}=2R$$.

$$GI \cdot IJ = 2R \sin(\angle PCB) \cdot CI$$.

Note that $$\angle PCB = \angle ICF$$ by the fourth of Euclid's common notions.

Now working with $$\sin(\angle ICF) \cdot CI$$ in $$\triangle CFI$$.

By the definition of sine, $$ \sin(\angle ICF)= \dfrac{\rho}{CI}$$, thus $$\sin(\angle ICF) \cdot CI = \rho$$.

Plugging this in to the earlier equation yields $$(R+d) \cdot (R-d) = 2R \cdot \rho \iff d^2=R(R-2 \rho) \,\!$$.