My solution: $X$ and $Y$ are the intersection points of $P_{1}Q_{2}$ and $P_{2}Q_{1}$ with $AC$ respectively. Using menelaus we get that $XA$=$YC$. So $M$ is the midpoint of $XY$. So $\frac{sin(\angle Q_1RM)}{sin(\angle P_1RM)}=\frac{sin(\angle RYC)}{sin(\angle RXA)}$. $\frac{sin(\angle Q_1YC)}{Q_1C}=\frac{sin(\angle YQ_1C)}{YC}$ and $\frac{sin(\angle P_1XA)}{P_1A}=\frac{sin(\angle XP_1A)}{XA}$. So $\frac{sin(\angle RYC)}{sin(\angle RXA)}=\frac{sin(\angle RQ_1Q_2)}{sin(\angle RP_1P_2)}.\frac{BQ_2}{BP_2}$ . $\frac{sin(\angle RP_1P_2)}{BQ_2}=\frac{sin(\angle ABC)}{P_1Q_2}$ and $\frac{sin(\angle RQ_1Q_2)}{BP_2}=\frac{sin(\angle ABC)}{P_2Q_1}$. So $\frac{sin(\angle Q_1RM)}{sin(\angle P_1RM)}=\frac{sin(\angle RYC)}{sin(\angle RXA)}=\frac{P_1Q_2}{P_2Q_1}$. $P_1P_2RS$ and $Q_1Q_2RS$ are cyclic. So $\angle SRQ_1=\angle P_2P_1S$ and $\angle SRP_1=\angle SP_2P_1$. We also know that $\triangle Q_1SP_2$ and $\triangle Q_2SP_1$ are similar. So $\frac{sin(\angle P_1RS)}{sin(\angle Q_1RS)}=\frac{P_1Q_2}{P_2Q_1}$. So $\frac{sin(\angle P_1RS)}{sin(\angle Q_1RS)}=\frac{sin(\angle Q_1RM)}{sin(\angle P_1RM)}$. From the Lemma at here http://www.artofproblemsolving.com/community/c6h1095783 , we get that the angles $P_1RS$ and $Q_1RM$ are equal.

My solution: Lemma. If we have: $S = \dfrac{\alpha_1A_1+\alpha_2A_2+ ... +\alpha_nA_n}{\alpha_1+\alpha_2+ ... +\alpha_n}$ Then for arbitrary points $X, Y$ we’ll have: $S_{SXY} = \dfrac{\alpha_1S_{A_1XY}+\alpha_2S_{A_2XY}+ ... +\alpha_nS_{A_nXY}}{\alpha_1+\alpha_2+ ... +\alpha_n}$ Proof. The proof which bases on the notice that $S_{ABC} = \frac{1}{2}.AB\wedge AC$ is very easy. Back to our main problem. $P_1Q_2, P_2Q_1\cap{AC} = X, Y$, resp. By Menelaus theorem: $\frac{\overline{AX}}{\overline{XC}} = -\frac{\overline{AP_1}}{\overline{P_1B}}.\frac{\overline{BQ_2}}{\overline{Q_2C}} = -\frac{\overline{BP_2}}{\overline{P_2A}}.\frac{\overline{CQ_1}}{\overline{Q_1B}} = \frac{\overline{CY}}{\overline{YA}}$ $\Rightarrow X, Y$ are symmetric WRT the midpoint $M$ of $AC$ $\Rightarrow$ Set $\frac{\overline{AX}}{\overline{XC}} = \frac{\overline{CY}}{\overline{YA}} = k$ $\Rightarrow X = \frac{A+k.C}{1+k}; Y = \frac{C+k.A}{1+k}$ $\Rightarrow S_{XP_2Q_1} = \frac{S_{AP_2Q_1}+k.S_{CP_2Q_1}}{1+k}; S_{YP_1Q_2} = \frac{S_{CP_1Q_2}+k.S_{AP_1Q_2}}{1+k}$ (lemma) (1) But: $S_{AP_2Q_1} = S_{P_1BQ_1} = S_{P_1Q_2C} (\because AP_2 = P_1B, BQ_1 = Q_2C)$ (2) And $S_{CP_2Q_1} = S_{Q_2P_2B} = S_{Q_2AP_1} (\because CQ_1 = Q_2B, P_2B = AP_1)$ (3) (1), (2), (3) $\Rightarrow S_{XP_2Q_1} = S_{YP_1Q_2}$ $\Rightarrow XR.P_2Q_1.sin\angle{R} = YR.P_1Q_2.sin\angle{R}$ $\Rightarrow \frac{RX}{RY} = \frac{P_1Q_2}{P_2Q_1}$ (4) On the other hand, $K, L$ are the projections of $S$ on $P_1Q_2, P_2 Q_1$ $\Rightarrow \frac{SK}{SL} = \frac{P_1Q_2}{P_2Q_1} (\because \triangle{SP_1Q_2} \sim \triangle{SP_2Q_1})$ (5) (4), (5) $\Rightarrow \frac{RX}{RY} = \frac{SK}{SL} \Rightarrow RS$ is the symmedian of $\triangle{RXY}$ $\Rightarrow RS, RM$ are isogonal in $\angle{P_1RQ_1}$ Q.E.D

Attachments:

My solution : Let $ B' $ be the reflection of $ B $ in the midpoint of $ P_1Q_1 $ . Since $ \{ P_1, P_2 \}, \{ Q_1, Q_2 \} $ are isotomic conjugate on $ BA, BC $, respectively , so $ \{ Y \equiv P_1Q_2 \cap AC , Z \equiv P_2Q_1 \cap AC \} $ are isotomic conjugate on $ AC \Longrightarrow M $ is the midpoint of $ YZ $ . From $ BQ_1 \parallel P_1B' , BQ_1 = P_1B' $ and $ BP_2=P_1A \Longrightarrow \triangle BP_2Q_1 $ and $ \triangle P_1AB' $ are congruent and homothetic . Similarly, $ \triangle BP_1Q_2 $ and $ \triangle Q_1B'C $ are congruent and homothetic $ \Longrightarrow \triangle RYZ \cup M $ and $ \triangle B'CA \cup M $ are homothetic . From the discussion above $ \Longrightarrow $ $ R, M, B' $ are collinear $ \Longrightarrow $ $ RM $ pass through the complement of $ B $ WRT $ \triangle RP_1Q_1 $ , so from Three concurrent radical axes $ RM $ is the isogonal conjugate of $ RS $ WRT $ \angle P_1RQ_1 $ . i.e. $ \angle P_1RS=\angle Q_1RM $ Q.E.D

Denote $Y \equiv P_1Q_2 \cap AC, Z \equiv P_2Q_1 \cap AC.$ By Menelaus' Theorem for $\triangle ABC$ cut by $\overline{P_1Q_2Y}$ and $\overline{P_2Q_1Z}$ we obtain \[1 = \frac{P_1A}{P_1B} \cdot \frac{Q_2B}{Q_2C} \cdot \frac{YC}{YA} = \frac{P_2B}{P_2A} \cdot \frac{Q_1C}{Q_1B} \cdot \frac{ZA}{ZC}.\]Since $\tfrac{P_1A}{P_1B} = \tfrac{P_2B}{P_2A}$ and $\tfrac{Q_2B}{Q_2C} = \tfrac{Q_1C}{Q_1B}$, it follows that $\tfrac{YC}{YA} = \tfrac{ZA}{ZC}.$ Hence, $Y$ and $Z$ are symmetric about $\overline{AC}$, so $M$ is the midpoint of $\overline{YZ}.$ Thus, $RM$ is the R-median of $\triangle RYZ$, so it is sufficient to show that $S$ lies on the R-symmedian of this triangle. We will equivalently (Characterization 2) show that $\text{dist}(S, RY) : \text{dist}(S, RZ) = RY : RZ.$ By the Law of Sines we have \begin{align*} \frac{RY}{RZ} = \frac{\sin\angle RZY}{\sin\angle RYZ} = \frac{\sin\angle Q_1ZC}{\sin\angle P_1YA} = \frac{Q_1C \cdot \frac{\sin\angle ZQ_1C}{ZC}}{P_1A \cdot \frac{\sin\angle YP_1A}{YA}} &= \frac{Q_1C}{\sin\angle YP_1A} \cdot \frac{\sin\angle ZQ_1C}{P_1A} \\ &= \frac{BQ_2}{\sin\angle BP_1Q_2} \cdot \frac{\sin\angle BQ_1P_2}{BP_2} \\ &= \frac{P_1Q_2}{\sin\angle P_1BQ_2} \cdot \frac{\sin\angle P_2BQ_1}{P_2Q_1} = \frac{P_1Q_2}{P_2Q_1}. \end{align*}To finish, note that $S$ is the center of spiral similarity that sends $\overline{P_1Q_2} \mapsto \overline{P_2Q_1}$, implying that $\triangle SP_1Q_2 \sim \triangle SP_2Q_1.$ Hence, \[\frac{\text{dist}(S, RY)}{\text{dist}(S, RZ)} = \frac{\text{dist}(S, P_1Q_2)}{\text{dist}(S, P_2Q_1)} = \frac{P_1Q_2}{P_2Q_1} = \frac{RY}{RZ},\]as desired. $\square$

Note that \[ [AP_1Q_2]=\frac{AP_1}{AB}\frac{BQ_2}{BC}[ABC]=\frac{BP_2}{AB}\frac{CQ_1}{BC}[ABC]=[CP_2Q_1] \]and similarly $[AP_2Q_1]=[CP_1Q_2]$. Thus \[ [MP_1Q_2]=\frac{1}{2}\left([AP_1Q_2]+[CP_1Q_2]\right)=\frac{1}{2}\left([CP_2Q_1]+[AP_2Q_1]\right)=[MP_2Q_1] \]But by spiral similarity, \[ \frac{d(S,P_1Q_2)}{d(S,P_2Q_1)}=\frac{P_1Q_2}{P_2Q_1}=\frac{d(M,P_2Q_1)}{d(M,P_1Q_2)} \]where $d(X,\ell)$ is the distance from point $X$ to line $\ell$, implying what we want.

Excellent angle-chase tutorial problem! ComplexPhi wrote: Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal. Let $L, N$ be midpoints of $BC$ and $BA$ respectively. Note that $S$ is the spiral center for $P_2P_1 \mapsto Q_1Q_2$; hence $S \in (BLN)$. Let $S_1, K$ be the other intersections of the line through $S$ parallel to $BC$ with $(BLN)$ and $(Q_1RQ_2)$, respectively. Claim: Points $R, K$ and $M$ are collinear. (Proof) Draw isosceles trapezoid $MCBM^{*}$. Note that $M^{*} \in (BLN)$ and $S_1M^{*}NS$ is also an isosceles trapezoid. Applying equality of arcs $S_1M^{*}$ and $SN$ we conclude $$\measuredangle SKM=\measuredangle S_1SM^{*}=\measuredangle SBN=\measuredangle SQ_1R=\measuredangle SKR,$$hence the claim is proven. $\blacksquare$ Note that $$\measuredangle MRQ_2=\measuredangle KRQ_2=\measuredangle SQ_2Q_1=\measuredangle SP_1P_2=\measuredangle SRP_2,$$hence lines $RS$ and $RM$ are isogonal in angle $P_2RQ_2$ which implies our required result. $\blacksquare$

Beautiful problem! One of my favourites ComplexPhi wrote: Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal. Let $T=P_1Q_2 \cap AC, L=P_2Q_1 \cap AC.$ Define $\alpha=\angle P_2P_1R, \beta=\angle RQ_1Q_2, x=\angle CLQ_1$ and $y=\angle P_1TA.$ Key Claim: $M$ is also the midpoint of $TL.$ Proof: Using Menelaus on $\triangle ABC$ with the transversals $P_1Q_2,P_2,Q_1$ give $$\frac{BQ_2}{Q_2C} \cdot \frac{CT}{TA} \cdot \frac{AP_1}{P_1B} =-1=\frac{CQ_1}{Q_1B} \cdot\frac{BP_2}{P_2A} \cdot\frac{AL}{LC}$$$$\implies TA=CL $$This yields the result as $M$ is the midpoint of $AC.$ $\square$ Hence, it suffices to show that $S$ lies on the $R\text{-symmedian}$ of $\triangle RTL.$ Let $\delta (U,VW)$ denote the distance from point $U$ to line $VW.$ Then since $\triangle SP_1Q_2 \sim \triangle SP_2Q_1$ by spiral similarity, we get \begin{align*} \frac{\delta(S,TR)}{\delta(S,LR)} &= \frac{P_1Q_2}{P_2Q_1} \\ &=\frac{BQ_2}{\sin \alpha} \cdot \frac{\sin \beta}{BP_2} \qquad \text{(by sine rule on } \triangle BP_2Q_1, \triangle BP_1Q_2) \\ &=\frac{CQ_1}{\sin \alpha} \cdot \frac{\sin \beta}{AP_1} \\ &=\frac{LC}{TA} \cdot \frac{\sin x}{\sin y} \qquad \text{(by sine rule on } \triangle P_1TA, \triangle Q_1CL) \\ &=\frac{RT}{RL} \qquad \text{(by sine rule on } \triangle RTL) \end{align*}Hence, we a well known lemma (for instance see Theorem 9.4 in Lemmas in olympiad Geometry) we get that $RS$ is the $R\text{-symmedian}$ of $\triangle RTL.$ $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; 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ComplexPhi wrote: Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal. Even better proof Let $O$ be the circumcenter of triangle $ABC$. Denote by $X$ and $Y$ the midpoints of $BA$ and $BC$. Let $Q, P$ be points on $\odot(Q_1RQ_2)$ and $\odot(P_1RP_2)$ such that $SQ \parallel BC$ and $SP \parallel BA$. Notice that $$\measuredangle QRQ_1=\measuredangle Q_2Q_1S=\measuredangle P_1P_2S=\measuredangle P_1RS$$so $RS, RQ$ are isogonal in the angle $P_1RQ_1$. Similarly, $RS, RP$ are also isogonal in angle $P_1RQ_1$. Note that $S$ is the spiral center $P_1P_2 \mapsto Q_2Q_1$ hence $X \mapsto Y$ so $S$ lies on $(OBXY)$. Since $Q_2SQQ_1$ and $P_1SPP_2$ are isosceles trapezoids, we see that $Q$ and $P$ are the reflections of $S$ in lines $OY$ and $OX$ respectively. Finally, $S$ lies on $\odot(OXY)$, so by Simson's line, $PQ$ passes through the orthocenter $M$ of $\triangle OXY$, proving the claim $\blacksquare$