59th Ukrainian National Mathematical Olympiad

First, we are going to prove that $\angle BDP=\angle AMN$ and $\angle PDC=ANM$. Indeed, $MP\parallel AC$, $NP\parallel AB$, because $PMAN$ is a parallelogram (fig. 38). Then $\angle MPB=\angle ABC=\angle ACB=\angle NPC$, meaning that $△BMP\sim △PNC$. We also note that $BC$ is a bisector of interior angle $△MPN$. Then $\frac{BP}{PC}=\frac{MP}{NC}$. Using $MP=AN$ one gets $NC=NP=AM\cdot \frac{PB}{PC}=\frac{AN}{AM}$.

Notice that $△BDC$ is isosceles, and $\angle CBD=\angle BCD$. By cosine law for triangles $BPD$ and $CPD$: $$\frac{BP}{\sin \angle BDP}=\frac{PD}{\sin \angle CBD}=\frac{PD}{\sin \angle BCD}=\frac{PC}{\sin \angle PDC}.$$ From here it follows that $\frac{\sin \angle PDC}{\sin \angle BDP}=\frac{PC}{BP}$ and $\frac{\sin \angle PDC}{\sin \angle BDP}=\frac{AM}{AN}$. By sine law for $△AMN$ we have $\frac{\sin \angle ANM}{\sin \angle AMN}=\frac{AM}{AN}$. Finally we get $\frac{\sin \angle PDC}{\sin \angle BDP}=\frac{\sin \angle ANM}{\sin \angle AMN}$.

Noticing $$\angle PDC+\angle BDP=\angle BDC=180^{\circ }-\angle MAN=\angle MNA+\angle AMN$$ yields $\angle PDC=\angle MNA$ and $\angle PDB=\angle AMN$.

Going further, $\angle QMD=\angle AMN=\angle QDB$, hence $MQBD$ is cyclic. Considering triangle $CPD$ and $PLM$ gives $$\angle CDP=\angle NMP=\angle ANM,\angle LPM=90^{\circ }-\angle MPB=90^{\circ }-\angle ACB=\angle PCD,$$ meaning that $△CPD\sim △PLM$.

Therefore $\frac{CP}{LP}=\frac{CD}{MP}$. Using $MP=MB$ and $CD=DB$, one gets $\frac{CP}{LP}=\frac{BD}{MB}$.

Considering triangles $LPC$ and $MBD$ gives $\angle MBD=\angle LPC=90^{\circ }$, implying (together with the previous equality) $△LPC\sim △MBD$.

Finally, $\angle LPC=\angle MDB=180^{\circ }-\angle BQM$, meaning that $\angle BQL+\angle LCB=180^{\circ }$, i.e. that the quadrilateral $BQLC$ is cyclic.