41st Balkan Mathematical Olympiad

From Menelaus theorem for the triangle $△ABC$ and the lines $MQ$, $KN$ and $PL$ we get $$\overline{AU}\overline{UB}=-\overline{AQ}\overline{QC}\cdot \overline{CM}\overline{MB},$$ $$\overline{BV}\overline{VC}=-\overline{BL}\overline{LA}\cdot \overline{AP}\overline{PC},$$ $$\overline{CW}\overline{WA}=-\overline{CN}\overline{NB}\cdot \overline{BK}\overline{KA}.$$ After multiplying them we get: $$\overline{AU}\cdot \overline{BV}\cdot \overline{CW}=-\overline{AQ}\cdot \overline{CM}\cdot \overline{BL}\cdot \overline{AP}\cdot \overline{CN}\cdot \overline{BK}=-1.$$ Hence by the converse Menelaus theorem the points $U,V$ and $W$ are collinear, implying $△ALP$ and $△RMN$ are coaxial. Now by Desargues theorem, $△ALP$ and $△RMN$ are copolar, hence $A,R$ and $D$ are collinear. Let $S$ be the intersection of $LP$ and $MQ$. Similarly $△BKN$ and $△SQP$ are coaxial, implying they are copolar, hence $B,S$ and $E$ are collinear. Now since $U,V$ and $W$ are collinear, $△ABC$ and $△RST$ are coaxial and by Desargues theorem they are copolar, hence $AR\equiv DR$, $BS\equiv BE$ and $CT$ are concurrent. The lines $AR,BS$ and $CT$ cannot be parallel as $R,S$ and $T$ are inside $△ABC$. $\square$