Vanish the camouflage

Just reflect the point of the interior say $P$ in $BC$ to obtain $P'$ .Thus , $\angle PBC = \angle P'BC \ and \ \angle PCB = \angle P'CB$ by the reflection properties.Thus we have:

$\angle BP'C = 180 - \angle P'BC - \angle P'CB = 180 - (\angle PBC + \angle PCB) = 180 - \angle BAC$

Since $\angle BP'C =180 - \angle BAC$ , we conclude that $\Box ABP'C$ is cyclic.Denote $R(ABC)$ as circumradius of $\Delta ABC$ . Thus , $R(ABC) = R(P'BC)$

Also $\Delta PBC \cong \Delta P'BC$ using ( $A-S-A$ test). Thus , $R(PBC) = R(P'BC)$

Thus finally , $R(ABC) = R(PBC) \Rightarrow \boxed{R_1 = R_2}$ .

Using sine law on triangle ABC:Equation 1* /(BC/ sin (<BAC) = 2R. Using sine law on triangle PBC: Equation 2* BC/sin (<BPC) = 2r. We know that <BAC=<PBC+<PCB and <BPC=180-(<PBC+<PCB). We also know the identity sin (x) = sin (180-x). So sin (<BAC)=sin (<BPC). So Equation 1 is equal to Equation 2, which means R=r