Let $l$ : $B$ on $l$ and $l // AC$. $l$ cross $(BQ_2P_1)$ at $B, A_1$ and $(BQ_1P_2)$ at $B, C_1$. $Q_2P_1$ cross $AC$ at $T_1$ and $P_2Q_1$ cross $AC$ at $T_2$. $M_1 and M_2$ are Miquel points for $(T_1Q_2, AB, AC, BC) and (P_2T_2, AB, BC, AC)$. Lemma $1$. $M_1T_1$ tangent $(T_1RT_2)$ $M_1P_1A \sim M_1Q_2C$ (because $M_1$ lie on $(ABC)$ and $(BQ_2P_1)$ so $\angle M_1P_1A = \angle M_1Q_2B$ and $\angle M_1AB = \angle M_1BC$). We have: $ \frac{M_1C}{M_1A}\ = \frac{CQ_2}{P_1A}\ = \frac{BQ_1}{BP_2} $ but else $\angle P_2BQ_1 = \angle AM_1C$ so $AM_1C \sim P_2BQ_1$. $\angle M_1AC = \angle M_1P_1T_1 = \angle P_1P_2R$ so$M_1P_1$ tangent $(P_1P_2R)$ so $\angle P_2RT_1 = \angle M_1P_1B = \angle M_1T_1C$. It means $M_1T_1$ tangent $(T_1RT_2)$. $\square$ Lemma $2$. $A_1C_1T_2T_1$ is concircle. $A_1B//AT_1$ so $M_1$ lie on $A_1T_1$ from theorem Reim. So $A_1T_1$ and $C_1T_2$ are tangent $(T_1RT_2)$ so $\angle A_1T_1T_2 = \angle C_1T_2T_1$, but also $AC//A_1C_1$. It means that $A_1C_1T_1T_2$ is a isosceles trapezois so it is concircle. $\square$ Lemma $3$. $A_1RST_1$ is concircle. $\angle T_1A_1S = \angle 180^{\circ} - \angle M_1P_1B - \angle BP_1S = \angle 180^{\circ} - \angle P_2RT_1 - \angle SPT_2$(because from lemma $1$ $M_1P_1$ tangent $(P_1P_2R)$) = $\angle SRT_1$. It means that $A_1RST_1$ is concircle. $\square$ Solution: From lemma $3$ $A_1RST_1$ and $C_1RST_2$ are concircles; from lemma $2$ $A_1C_1T_2T_1$ is concircle. So from radixal axes $A_1T_1, C_1T_2, RS$ intersect at one point. But from lemma$2$ $A_1T_1$ and $C_1T_2$ are tangent $(T_1RT_2)$ so $RS$ is symmedian in triangle $T_1RT_2$ and $RM$ is median(from theorem Menelay easy to show $AT_1 = CT_2$). Finaly, $\angle P_1RS = \angle Q_1RM$. $\square$

Thanks.Nice!