circlesgeometrytriangles
Prove that, if $d_a$, $d_b$, $d_c$ are distances between the center of circumscribed circle and acute triangle sides ($a$, $b$, $c$ accordingly), then
$$d_a+d_b+d_c=R+r$$
I didn't do much, because I can not quite think of anything useful except somehow using this equations
$$d_aa+d_bb+d_cc=(a+b+c)\cdot r= 2S=2\sqrt{p(p-a)(p-b)(p-c)}$$
$$R=\frac{abc}{4S}$$
Where $S$ is area, $r$ is inradius, $R$ is circumradius, and $p$ is perimeter divided by two.
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, $2Rr_{A}=DI_{A}\cdot I_{A}A=OI_{A}^{2}-R^{2}$
Accidentally, I've found this open-access reference:
Yiu, Paul. The Congruent-Incircle Cevians of a Triangle. Missouri J. Math. Sci. 15 (2003), no. 1, 21--32. doi:10.35834/2003/1501021. https://projecteuclid.org/euclid.mjms/1567216820
For such cevians they use a term "the congruent-incircle cevians of a triangle".
Best Answer
For the acute triangle $ABC$ with the circumcenter $O$
\begin{align} \angle CAB&=\alpha ,\quad \angle ABC=\beta ,\quad \angle BCA=\gamma ,\\ \angle COB&=2\alpha ,\quad \angle AOC=2\beta ,\quad \angle BOA=2\gamma . \end{align}
Triangles $COB$, $AOC$ and $BOA$ are isosceles, hence
\begin{align} \angle COD&=\alpha ,\quad \angle AOE=\beta ,\quad \angle BOF=\gamma \end{align}
and \begin{align} d_a=|OD|&=R\cos\alpha ,\\ d_b=|OE|&=R\cos\beta ,\\ d_c=|OF|&=R\cos\gamma ,\\ d_a+d_b+d_c&= R(\cos\alpha+\cos\beta+\cos\gamma) . \end{align}
And it is well-known that for any triangle
\begin{align} \cos\alpha+\cos\beta+\cos\gamma&=\frac rR+1 ,\\ \text{so, }\quad d_a+d_b+d_c&= R\cdot\left(\frac rR+1\right) =r+R . \end{align}