THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

a) Let $S$ be the reflection of $C$ across the line $EF$. As $AE=AF$, it follows that $\angle AFE=\angle AEF$, hence $\angle BFE=\angle FEC=\angle FES$, therefore $FB$ and $ES$ are parallel. $FBES$ is a trapezoid (or a parallelogram); let $M$ be the intersection point of its diagonals. According to the billiards problem, $M$ is the point of the line $EF$ for which the sum $BM+MC$ is minimum. Triangles $BFM$ and $SEM$ are similar, therefore $\frac{FM}{EM}=\frac{FB}{ES}=\frac{FB}{EC}=\frac{s-b}{s-c}$. It follows that $\frac{\sin (FAM)}{\sin (EAM)}=\frac{[FAM]}{[EAM]}=\frac{FM}{EM}=\frac{s-b}{s-c}$. Multiplying this with the two analogous relations we get $$\frac{\sin (FAM)}{\sin (EAM)}\cdot \frac{\sin (ECP)}{\sin (DCP)}\cdot \frac{\sin (DBN)}{\sin (NBF)}=1.$$ By the converse of the trigonometric form of Ceva's theorem, it follows that the lines $AM$, $BN$, and $CP$ are concurrent.

b) We have $\frac{FM}{EM}=\frac{p-b}{p-c}=\frac{BD}{CD}=\frac{BF}{CE}$, hence, from a well-known problem ("The gliding bisector Theorem") it follows that $DM$ is parallel to the bisector of angle $\angle BAC$. The bisector is perpendicular to $EF$, therefore $DM$ is also perpendicular to $EF$.