Throughout the solution, $[XYZ]$ and $d(X, YZ)$ denote the area of the triangle $XYZ$ and the distance from the point $X$ to the line $YZ$, respectively.

Since the quadrilaterals $SRQ_2Q_1$ and $SRP_2P_1$ are cyclic, $\angle SQ_1R = \angle SQ_2R$ and $\angle SP_1R = \angle SP_2R$ (see Fig. 6), so the triangles $SP_1Q_2$ and $SP_2Q_1$ are similar, and
$$\frac{d(S, P_1Q_2)}{d(S, P_2Q_1)} = \frac{P_1Q_2}{P_2Q_1}. \qquad (1)$$
Let $K$ and $L$ be the midpoints of $AB$ and $AC$, respectively; then $P_1K = P_2K$ and $Q_1L = Q_2L$. Therefore, $d(Q_1, MP_1) + d(Q_2, MP_1) = 2d(L, MP_1)$, so
$$[MP_1Q_2] + [MP_1Q_1] = 2[MP_1L] = ML \cdot d(P_1, ML) = [ABC]/2.$$
Similarly, $[MP_2Q_1] + [MP_1Q_1] = [ABC]/2 = [MP_1Q_2] + [MP_1Q_1]$. Thus $[MP_1Q_2] = [MP_2Q_1]$, whence
$$\frac{d(M, P_1Q_2)}{d(M, P_2Q_1)} = \frac{P_2Q_1}{P_1Q_2}. \qquad (2)$$
In the angle $P_1RQ_1$, the condition $d(X, P_1Q_2)/d(X, P_2Q_1) = \alpha$ determines a ray emanating from $R$. Moreover, rays symmetric with respect to the bisector of $\angle P_1RQ_1$ correspond to reciprocal values of $\alpha$. Consequently, relations (1) and (2) show that the rays $RS$ and $RM$ are symmetric with respect to this angle bisector, and the conclusion follows.