Question 28
Let $A' \in (BC)$, $B' \in (AC)$, $C' \in (AB)$ be the points of tangency of the excribed circles of the triangle $ABC$ with the sides of $ABC$. Let $R'$ the circumradius of $A'B'C'$. Show that
$$
R' = \frac{1}{2r} \sqrt{2R(2R - h_a)(2R - h_b)(2R - h_c)},
$$
where, as usual, $R$ is the circumradius of $ABC$, $r$ is the inradius of $ABC$, and $h_a, h_b, h_c$ are the lengths of the altitudes of $ABC$.
$$
R' = \frac{1}{2r} \sqrt{2R(2R - h_a)(2R - h_b)(2R - h_c)},
$$
where, as usual, $R$ is the circumradius of $ABC$, $r$ is the inradius of $ABC$, and $h_a, h_b, h_c$ are the lengths of the altitudes of $ABC$.