Consider the triangle $SLM$. $LC$ and $MC$ are the tangent lines to the circumscribed circle of this triangle drawn at the points $L$ and $M$, therefore $SC$ is a simedian, and so $\frac{MQ}{CL} = \frac{MS^2}{SL^2}$.

Similarly, $\frac{KP}{PM} = \frac{KS^2}{SM^2}$. Let $E$ be the point of intersection of the lines $KL$ and $MR$. Since the lines $ME$, $KQ$ and $LP$ are concurrent, by the Ceva's theorem we have that $\frac{KP}{PM} \cdot \frac{MQ}{QL} \cdot \frac{LE}{EK} = 1$, hence applying the above equalities we obtain that $\frac{LE}{EK} = \frac{SL^2}{KS^2}$. This implies that the line $SE$ is a simedian of the triangle $SKL$, therefore it passes through the point of intersection of the tangents to the circumscribed circle of the triangle $KSL$ that are drawn at the points $L$ and $K$, that is through the point $B$ (Fig. 49). So, we have proved that the points $B$, $S$, $R$ and $M$ are collinear.

Next we will prove that the point $T$ also belongs to this line. By the Ceva's theorem, it is sufficient to prove that $\frac{AP}{\sin \angle AMP} = \frac{AM}{\sin \angle APM}$, $\frac{PS}{\sin \angle KMB} = \frac{MS}{\sin \angle SPM}$. Dividing the last two inequalities we obtain that $\frac{AP}{PS} = \frac{AM \sin \angle AMK}{MS \sin \angle KMB}$. Similarly, $\frac{CQ}{QS} = \frac{MC \sin \angle CML}{MS \sin \angle LMB}$. Substitute the last two relations into the equality that we need to prove:
$$\frac{AP}{PS} \cdot \frac{SQ}{QC} \cdot \frac{CM}{MA} = \frac{AP}{PS} = \frac{AM \sin \angle AMK}{MS \sin \angle KMB} \cdot \frac{MS \sin \angle LMB}{MC \sin \angle CML} \cdot \frac{CM}{MA} = \frac{\sin \angle AMK \cdot \sin \angle LMB}{\sin \angle KMB \cdot \sin \angle CML}$$
By the sine theorem for the triangles $BLM$ and $BMK$, we obtain:
$$\frac{BL}{BM} = \frac{\sin \angle LMB}{\sin \angle BLM} = \frac{\sin \angle LMB}{\sin \angle MLC} = \frac{\sin \angle LMB}{\sin \angle CML}, \text{ since the triangle MLC is isosceles. Similarly, } \frac{BK}{BM} = \frac{\sin \angle KMB}{\sin \angle AMK}$$
Therefore,
$$\frac{AP}{PS} \cdot \frac{SQ}{QC} \cdot \frac{CM}{MA} = \frac{\sin \angle AMK \cdot \sin \angle LMB}{\sin \angle KMB \cdot \sin \angle CML} = \frac{BM}{BK} \cdot \frac{BL}{BM} = 1,$$