Croatian Mathematical Society Competitions

The triangle $ABC$ is acute-angled, so the points $D$, $E$ and $F$ lie on the sides $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$, respectively. Since $|AB| > |AC|$, we have $|BD| > |CD|$.

Furthermore, from $\angle ACB < \angle ABC$ it follows that $C$ lies between $B$ and $P$. Also, $C$ lies between $A$ and $Q$, while $R$ lies on the segment $\overline{BF}$.

Let $H$ be the orthocentre of the triangle $ABC$, and let $M$ be the midpoint of its side $\overline{BC}$. The lines $DR$ and $EF$ are parallel, hence $\angle DRF = \angle EFA$. The quadrilateral $AFHE$ is cyclic ($\angle AFH = \angle AEH = 90^\circ$), so we have $\angle EFA = \angle EHA$. The quadrilateral $BDHF$ is cyclic as well ($\angle BDH = \angle BFH = 90^\circ$), hence $\angle EHA = \angle DHB = \angle DFB = \angle DFR$. Now from $\angle DRF = \angle DFR$ it follows that $|DR| = |DF|$.

Similarly, the lines $DQ$ and $FE$ are parallel, hence $\angle DQE = \angle FEA$. The quadrilateral $AFHE$ is cyclic, so we have $\angle FEA = \angle FHA$. The quadrilateral $CEHD$ is cyclic as well ($\angle CEH = \angle CDH = 90^\circ$), hence $\angle FHA = \angle DHC = \angle DEC = \angle DEQ$. Now from $\angle DQE = \angle DEQ$ it follows that $|DQ| = |DE|$.

The points $M$, $F$, $E$ and $D$ lie on the same circle (called the nine-point circle or the Feuerbach's circle), hence $\angle FMD = 180^\circ - \angle FED = \angle PED$. Also, $\angle FDH = \angle FBH = \angle ABE = 90^\circ - \angle CAB = \angle ACF = \angle ECH = \angle EDH$.

Therefore, $\angle FDM = 90^\circ - \angle FDH = 90^\circ - \angle EDH = \angle PDE$, so the triangles $FDM$ and $PDE$ are similar. Now we conclude that
$$\frac{|DF|}{|DM|} = \frac{|DP|}{|DE|},$$
from which it follows that $|DF| \cdot |DE| = |DP| \cdot |DM|$, i.e.
$$|DR| \cdot |DQ| = |DP| \cdot |DM|,$$
meaning that the quadrilateral $PQMR$ is cyclic.

Let $k$ be its circumscribed circle. Since $\angle NQP + \angle NRP < 180^\circ$, $N$ lies inside the circle $k$, i.e. on the segment $\overline{CM}$, and $N \neq M$ holds. Since $M$ is the midpoint of the segment $\overline{BC}$, we finally get $|BN| > |CN|$.