5. (5 points) As shown in the figure, in $\triangle A B C$, point $E$ is on $A B$, point $F$ is on $A C$, and $B F$ intersects $C E$ at point $P$. If $S$ quadrilateral $A E P F = S_{\triangle B E P} = S_{\triangle C F P} = 4$, then the area of $S_{\triangle B P C}$ is $\qquad$ .
【Solution】Solution: Connect $E F, A P$. According to the problem, it is not difficult to derive that the areas of $\triangle C E F$ and $\triangle B E F$ are equal and they share the same base, so their heights on base $E F$ are also equal. From this, we can conclude: $E F / / B C$, then: $C F: A F=B E: A E$; and $C F: A F=S_{\triangle C F P}: S_{\triangle A F P} ; B E: A E=S_{\triangle B E P}: S_{\triangle A E P}$; thus: $S_{\triangle C F P}: S_{\triangle A F P}=S_{\triangle B E P}: S_{\triangle A E P}$; also, since $S_{\triangle C F P}=S_{\triangle B E P}=4$; we can conclude that $A P$ bisects quadrilateral $A E P F$, i.e., $S_{\triangle A F P}=S_{\triangle A E P}=2$; therefore, we have: $A F: F C=1: 2$, so $S_{\triangle B A F}: S_{\triangle B F C}=1: 2$, thus the area of $\triangle B P C$ is: $4 \times 2 \times 2-4=12$, hence the answer is: 12.