Let $G$, $H$, $O$ be the centroid, orthocenter, and circumcenter of $\triangle ABC$. The homothety $\mathcal{H}(G, -\tfrac{1}{2})$ takes the excentral triangle to the triangle $\mathcal{T}$ formed by $r_A$, $r_B$, and $r_C$. Thus it takes the circumcenter of the excentral triangle (well-known to be $2O - I$; interpret $\triangle ABC$ as the orthic triangle of $\mathcal{T}$) to \[-\tfrac{1}{2}(2O - I - G) + G = -O + \tfrac{1}{2} I + \tfrac{3}{2}G = \tfrac{1}{2}(H + I)\](by the Euler line), so the desired circumcenter is the midpoint of $\overline{HI}$.

Construct the medial triangle of $ABC$, $DEF$, with $D, E, F$ the midpoints of $BC, CA, AB$. Note the angle bisector of $\angle BAC$ is parallel to the angle bisector of $\angle EDF$. Thus, the triangle formed by $r_A, r_B, r_C$ is the excentral triangle of the medial triangle. Let $S$, $N$ denote the incenter and circumcenter of the medial triangle. Then $S$ is the orthocenter of the triangle formed by $r_A, r_B, r_C$ with $N$ the Nine-Point Center of the same triangle, so the reflection of $N$ across $S$, $N'$ is the circumcenter of this triangle. Also, $H$ is the reflection of $O$, the circumcenter of $ABC$, about $N$. Thus $HN'$ is parallel to $OS$, and $HN' = OS$. Now consider a homothety about $G$, the centroid of $ABC$, of factor $-2$. $O$ is mapped to $H$. Since this maps the medial triangle $DEF$ to $ABC$, $S$, the incenter, maps to the incenter $I$ of $ABC$. Then $HI$ is parallel to $OS$, so it follows that $H, I, N'$ are collinear. $HN' = OS$ from before, and $HI$ = $2OS$, so it follows that $N'$ is the midpoint of $HI$, as desired.

Complex bash : Lets use the angular coefficient formula. We have that the angle bissector of A is perpendicular to $r_a$, the angle bissector of B is perpendicular to $r_b$ and the angle bissector of C is perpendicular to $r_c$. The angular coefficient formula of the line AI, with A being a vertice of A and I the incenter. Let $A=a^2$, $B=b^2$ and $C=c^2$ , the midpoint of arc BC is $-bc$ . Givving us $\frac{a^2 + bc}{\frac{1}{a^2} + \frac{1}{bc}}$ = $a^2 bc$. Let $X_a$ being the meeting point of $r_b$ and $r_c$. So we get $\frac{x_a - m_b}{\overline{x_a} - \overline{m_b}} = -a b^2 c $ and $\frac{x_a - m_c}{\overline{x_a} - \overline{m_c}} = -a b c^2 $ . Giving us $x_a - \frac{a^2 + c^2}{2} = \overline{x_a} -ab^2c + \frac{b^2(a^2 + c^2)}{2ac}$ and $x_a - \frac{a^2 + b^2}{2} = \overline{x_a} -ab c^2 + \frac{ac^2 +b^2c}{2ab} $ . $c x_a$ = $\frac{a^2 b^2 c + b^2 c^3}{2ac} + \frac{ a^2 c + c^3}{2} - \overline{x_a}ab^2c^2 $ $b x_a$ = $\frac{a^2 b^ 2 + b^2 c^2}{2ab} + \frac{ab^2 + b^3}{2} - \overline{x_a}ab^ 2 c^ 2 $ subtracting the two equations we get : $x_a$ = $\frac{a^3 c + ac ^ 3 - a ^ 3b - ab ^ 3 + a^2 b^ 2 - a^ 2c^ 2}{2a(c-b)}$. Picking the modulus formula, we get : $| x_a - \frac{[a^2 + b^2 + c^2 - ab - ac - bc]}{2} | $ $ | \frac{2a b^ 2 c + 2ab c^2}{ 2a ( c-b)} | $ $| \frac{2a(-b^2c + bc^2)}{2a(c-b)}|$ = $|\frac{bc(-b+c)}{(c-b)}|$ = $|-bc|$ = 1 . Using the annalog thing for $X_b$ and $X_c$ we get that both are 1 so the circuncenter of $x_a x_b x_c$ it's the midpoint of $HI$.

Let $M_a$ be the midpoint of arc $BC$ not containing $A$. Define $M_b$ and $M_c$ similarly. We also say $X$, $Y$, $Z$ are the vertices of the triangle formed and $R$ and $T$ are it's circumcenter and orthocenter, respectively. Claim: $\triangle ABC$, $\triangle M_a M_b M_c$ and $\triangle XYZ$ are all similar. Furthermore, there's a homotethy that takes $\triangle M_a M_b M_c \to \triangle XYZ$. It's known that $\triangle ABC \equiv \triangle M_a M_b M_c$. To prove they're all similar, just note that sides of $\triangle M_a M_b M_c$ and $\triangle XYZ$ are perpendicular to the angle bisectors of $\triangle ABC$. This means that their sides are parallel and therefore they're homothetic. Claim: In fact, all triangles are congruent. By the homothethy, the altitudes of $XYZ$ must be parallel to the angle bisectors of $\triangle ABC$. This eventually means that the orthic triangle of $XYZ$ is the medial triangle of $ABC$. So, they have the same $9$-point circle and, hence, the same radius. With all of this, we can finally finish the problem. As they share the same $9$-point circle, $HROT$ must be a parallelogram. But, by the homothety with ratio $-1$ talking $\triangle M_aM_bM_c \to \triangle XYZ$, $RIOT$ is also a parallogram. This also imples that $HRI$ are collinear. Now, we have $HR = OT = RI$, and we're done.