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 $EF$.