18.17. Let $M$ and $N$ be the midpoints of sides $CD$ and $DE$ of a regular hexagon $ABCDEF$, and $P$ be the point of intersection of segments $AM$ and $BN$.

a) Find the measure of the angle between the lines $AM$ and $BN$.

b) Prove that $S_{ABP} = S_{MDNP}$.
18.17. When rotating by $60^{\circ}$ around the center of this hexagon, which maps vertex $A$ to $B$, segment $C D$ transitions to $D E$, so point $M$ transitions to $N$. Thus, under this rotation, segment $A M$ transitions to $B N$, meaning the angle between these segments is $60^{\circ}$. Additionally, under this rotation, pentagon $A M D E F$ transitions to $B N E F A$, meaning their areas are equal. By cutting out the common part of these equal-area pentagons, pentagon $A P N E F$, we obtain two equal-area figures: triangle $A B P$ and quadrilateral $M D N P$.