(i) Let the line that passes $U$ and is parallel to $AC$ intersect $CR$ at $X$, and let the line that passes $U$ and is parallel to $AB$ intersect $BS$ at $Y$. Since $UY \parallel AB$, $\triangle PUT \sim \triangle PAB$. From this we get $\frac{PU}{AP} = \frac{PY}{BP}$. Similarly $\frac{PU}{AP} = \frac{PX}{CP}$. Hence $\frac{PY}{BP} = \frac{PX}{CP}$, and therefore $XY \parallel BC$.

(ii) Since $\angle VWP = \angle XRS = \angle PBC = \angle BYX$, it follows that $R$, $S$, $X$, $Y$ are concyclic, and also $V$, $W$, $X$, $Y$ are concyclic.

(iii) Since $V$, $W$, $X$, $Y$ are concyclic, $\angle BYU = \angle CYU$, and from this we obtain $\angle ABP = \angle ACP$ (because $UY \parallel AB$, $UX \parallel AC$).

(iv) Let $AP$ meet $CQ$ at $D$, let $BP$ meet $AC$ at $E$, and let $CP$ meet $AB$ at $F$. To prove $\angle CAP = \angle BAQ$, it suffices to prove $\triangle BAQ \sim \triangle CAD \sim \triangle EAP$.
Since $BPCQ$ is a parallelogram, and $\angle ABP = \angle ACP$, $\angle ABQ = \angle ACD = \angle AEP$, and also $PC = BQ$. So if $\frac{BQ}{AB} = \frac{EP}{AE}$ or $\frac{PC}{AB} = \frac{EP}{AE}$, then we can obtain $\triangle BAQ \sim \triangle EAP$.

(v) By the Law of Sines, $\frac{AB}{AE} = \frac{\sin \angle AEP}{\sin \angle ABP}$, $\frac{PC}{PE} = \frac{\sin \angle PEC}{\sin \angle ACP}$. Since $\angle ABP = \angle ACP$ and $\angle PEC$, $\angle AEP$ are supplementary, we get $\frac{AB}{AE} = \frac{PC}{PE}$. This completes the proof!