Problem 1862

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 $\triangle BMP \sim \triangle PNC$. We also note that $BC$ is a bisector of interior angle $\triangle MPN$. Then $\frac{BP}{PC} = \frac{MP}{NC}$. Using $MP = AN$ one gets $NC = NP = AM \cdot \frac{PB}{PC} = \frac{AN}{AM}$.

Notice that $\triangle BDC$ 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 $\triangle AMN$
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
$$\angle CDP = \angle NMP = \angle ANM, \quad \angle LPM = 90^\circ - \angle MPB = 90^\circ - \angle ACB = \angle PCD,$$
meaning that $\triangle CPD \sim \triangle PLM$.

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) $\triangle LPC \sim \triangle MBD$.

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.