Euler Triangle Formula

Theorem
Let $$d$$ be the distance between the incenter and the circumcenter of some triangle.

Then $$d^2 = R \left({R-2 \rho}\right)$$, where $$R$$ is the circumradius and $$\rho$$ is the inradius.

Lemma
Let the bisector of one angle of the triangle be produced to one point of the circumcircle $$P$$.

Then the distance of any of the vertices, of the triangle, left are equal to $$IP$$, where $$I$$ is the incenter of the triangle.

Proof (Lemma)

 * [[Image:Te1.PNG]]

In the figure we have that $$CP$$ is the bisector of $$\angle ACB$$.

Now how $$\angle ABP$$ and $$\angle ACP$$ subtend the same arc then:
 * $$\angle ACP = \angle ABP = \dfrac{\angle C}{2}$$

As $$IB$$ is the bisector of $$B$$ then:
 * $$\angle IBA = \dfrac{\angle B}{2} \implies \angle IBP = \dfrac{\angle B + \angle C}{2}$$

and in $$\triangle CIB$$ the supplement of $$\angle CIB = \dfrac{\angle C}{2} + \dfrac{\angle B}{2}$$.

Therefore $$\angle PIB = \angle IBP \implies IP = BP$$.

Proof (Theorem)

 * [[Image:Incenter Circumcenter Distance.png]]

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

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 $$(R+d) \cdot (R-d) = 2R \cdot \rho \iff d^2=R(R-2 \rho) \,\!$$.