Fig. 40

Denote $AC = b$, $AB = c$, $\angle BAC = \alpha$, and $AE = AP = u$. As $BCEN$ is a parallelogram, $BN = CE = b-u$ and $BN \parallel CE$. The latter implies $\angle NBP = \alpha$ (Fig. 40). As $\angle QAE = \angle QBN$ and $\angle AQE = \angle BQN$, triangles $AQE$ and $BQN$ are similar.

a) The triangles $AEP$ and $BNP$ have areas $\frac{1}{2}u^2 \sin \alpha$ and $\frac{1}{2}(b-u)(c-u) \sin \alpha$, respectively. Thus $u^2 = (b-u)(c-u)$, whence $\frac{c-u}{u} = \frac{u}{b-u}$. Similarity of triangles $AQE$ and $BQN$ implies $\frac{AQ}{BQ} = \frac{AE}{BN}$ which, after defining $BQ = x$, rewrites to $\frac{c-x}{x} = \frac{u}{b-u}$. This equality reduces to a linear equation of $x$, meaning that it has only one root. By equality $\frac{c-u}{u} = \frac{u}{b-u}$, $x = u$ must be the only root. Thus $BQ = u = AP$, whence the midpoint of the side $AB$ coincides with the midpoint of the line segment $PQ$.

Fig. 41

b) Let $F$ be the midpoint of the side $AB$ and let $D$ be the point of intersection of the bisector of angle $A$ with side $BC$ (Fig. 41). By angle bisector theorem, $\frac{BD}{CD} = \frac{AB}{AC}$. Since $NE$ and $BC$ are parallel, triangles $ABC$ and $AQE$ are similar, whence also triangles $ABC$ and $BQN$ are similar. This and part a) of the problem together imply $\frac{CE}{EA} = \frac{BN}{BQ} = \frac{AC}{AB}$. Consequently, $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1 \cdot \frac{AB}{AC} \cdot \frac{AC}{AB} = 1$, giving the desired result by Ceva's theorem.