Question 47
In the triangle $ABC$, in which $\angle A = 60^\circ$, $D \in (BC)$ is such that $AD$ is the internal bisector of angle $\angle A$. Let $r_B, r_C$ and $r$ be, respectively, the inradii of the triangles $ABD$, $ADC$ and $ABC$. Show that $\frac{1}{r_B} + \frac{1}{r_C} = 2\left(\frac{1}{r} + \frac{1}{b} + \frac{1}{c}\right)$, where $b$ and $c$ are the lengths of the sides $AC$ and $AB$ of the triangle $ABC$.