Problems of Ukrainian Authors

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 $\frac{MQ}{CL}=\frac{M{S}^2}{S{L}^2}$.

Similarly, $\frac{KP}{PM}=\frac{K{S}^2}{S{M}^2}$. 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 $\frac{LE}{EK}=\frac{S{L}^2}{K{S}^2}$. 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, $$\frac{AP}{PS}\cdot \frac{SQ}{QC}\cdot \frac{CM}{MA}=\frac{\sin \angle AMK\cdot \sin \angle LMB}{\sin \angle KMB\cdot \sin \angle CML}=\frac{BM}{BK}\cdot \frac{BL}{BM}=1,$$