THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

From $△BMC\sim △PBC$ follows $\frac{BM}{PB}=\frac{BC}{PC}$, hence $CP=2BP$, so $[BP]=[PN]=[NC]$.



Then $\Delta NCD\equiv \Delta PBC$ (SAS) leads to $\widehat{DNC}=\widehat{BPC}$, that is $DN\perp CP$, so $[DN]$ is a median and an altitude in triangle $DPC$. Therefore triangle $DPC$ is isosceles, with $[DP]=[DC]$.

Denote $E$ the common point of the straight lines $CM$ and $AD$. Then $A$ is the midpoint of $[DE]$, hence $[NA]$ is a median in the right triangle $NDE$. This leads to $[NA]=[AD]$, hence triangle $ADN$ is isosceles. Since $AQ$ is a bisector, it is also an altitude, therefore $AQ\perp DN$. From $DN\perp CM$ follows $AQ\parallel MN$ (1).

Isosceles triangles $ANE$ and $DPC$ yield $\widehat{ANP}+\widehat{DPN}=\widehat{DEC}+\widehat{DCE}=90^{\circ }$, whence $DP\perp AN$. Since $AQ\perp DN$, the point $Q$ is the orthocenter of the triangle $ADN$, so $NQ\perp AD$.

This shows that $NQ\parallel AM$ and, using (1), we get that $AMNQ$ is a parallelogram. So the segments $[AM]$ and $[NQ]$ are parallel and congruent, hence $[BM]$ and $[NQ]$ are parallel and congruent, therefore $BMQN$ is a parallelogram.