Prove the distances between the circumcenter, $O$, and the three excenters, $I_1$, $I_2$, $I_3$, of a triangle are given by $$\begin{align} |OI_1|^2 &= R(R + 2r_1) \\ |OI_2|^2 &= R(R + 2r_2) \\ |OI_3|^2 &= R(R + 2r_3) \end{align}$$ where $R$ is the circumradius and $r_1$, $r_2$, $r_3$ are the respective exradii.

I know that the circumcenter is the point of concurrence of perpendicular bisectors of sides; center of circumscribed circle). I also know that excenter is the center of a circle that is tangent to the three lines extended along the sides of a triangle.


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angle bisector $AI$ cut the circumcircle of $\triangle ABC$ at $D$
$\angle DBI=\angle DBC+\angle IBC=\angle DAB+\angle ABI=\angle BID$ and then $DB=DI$
Likewise, $DC=DI$ and then $DB=BI=DC$
$I_{A}C$ bisect $\angle BCT$ $\Longrightarrow$ $\angle ICI_{A}=90^{\circ}$
thus, $DI=DC=DI_{A}$

the perpendicular bisector of $BC$ cut the circumcircle of $\triangle ABC$ at $M$
$\triangle SAI_{A}$ is similar to $\triangle BMD$
power of $I_{A}$ with respect to the circumcircle of $\triangle ABC$ is $OI_{A}^{2}-R^{2}$
Also, it is $I_{A}D\cdot I_{A}A$

$\triangle SAI_{A}\sim\triangle BMD$ $\Longrightarrow$ $\dfrac{MD}{BD}=\dfrac{I_{A}A}{SI_{A}}$
$\Longrightarrow$ $2Rr_{A}=BD\cdot I_{A}A=DI_{A}\cdot I_{A}A$

hence, $2Rr_{A}=DI_{A}\cdot I_{A}A=OI_{A}^{2}-R^{2}$


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