Let $I \equiv BP \cap CQ$ be the incenter of $\triangle ABC.$ $(U,R_A)$ is circumcircle of $\triangle APQ.$ Since $\angle PIQ=120^{\circ},$ then $I$ is midpoint of the arc $PQ$ of $(U)$ $\Longrightarrow$ $M \equiv IU \cap PQ$ is midpoint of $PQ.$ Let $J$ be the incenter of $\triangle APQ.$ $E$ is the projection of $I$ on $AC$ and $N$ lies on $\overline{IE},$ such that $IM=IN.$ $IMEP$ is cyclic due to the right angles at $M,E.$ Thus $\angle IEM=\angle IPQ=\angle IAQ$ $\Longrightarrow$ $\triangle IEM \sim \triangle IAQ$ $\Longrightarrow$ $\frac{_{IE}}{^{IN}}=\frac{_{IA}}{^{IQ}}=\frac{_{IA}}{^{IJ}}$ $\Longrightarrow$ $JN \parallel AC$ $\Longrightarrow$ $NE=r_2=IE-IN=r_1-\frac{_1}{^2}R_A.$