$\frac{[PEL]}{[PFN]}=\frac{PE \cdot PL}{PF \cdot PN} \ , \ \frac{[PFM]}{[PDL]}=\frac{PF \cdot PM}{PD \cdot PL} \ , \ \frac{[PDN]}{[PEM]}=\frac{PD \cdot PN}{PE \cdot PM} \Longrightarrow$ $\frac{[PEL]}{[PFN]} \cdot \frac{[PFM]}{[PDL]} \cdot \frac{[PDN]}{[PEM]}=\frac{PE \cdot PL}{PF \cdot PN} \cdot \frac{PF \cdot PM}{PD \cdot PL} \cdot \frac{PD \cdot PN}{PE \cdot PM}=1.$ But from paralellograms $PDBL,$ $PECM,$ $PFAN,$ we have $[PFN]=\tfrac{1}{2}[PFAN],$ $[PDL]=\tfrac{1}{2}[PDBL]$ and $[PEM]=\tfrac{1}{2}[PECM].$ Thus, the latter expression becomes $[PDBL] \cdot [PECM] \cdot [PFAN]=8 \cdot [PFM] \cdot [PEL] \cdot [PDN].$

We know ANPF and BDPL and CMPE are parallelogram so what we need is: [PNF] . [PDL] . [PEM] = [NPD] . [LPE] . [MPF] (Note that [ANPF] = 2[PNF] and ...) [NPD] / [PEM] = NP/PE . DP/PM [MPF] / [PDL] = FP/PL . MP/PD [LPE] / [PNF] = PE/EN . PL/PF so [NPD] / [PEM] . [MPF] / [PDL] . [LPE] / [PNF] = 1 ---> [PNF] . [PDL] . [PEM] = [NPD] . [LPE] . [MPF]

Let $P$ be a point inside a triangle $ABC$. A line through $P$ parallel to $AB$ meets $BC$ and $CA$ at points $L$ and $F$, respectively. A line through $P$ parallel to $BC$ meets $CA$ and $BA$ at points $M$ and $D$ respectively, and a line through $P$ parallel to $CA$ meets $AB$ and $BC$ at points $N$ and $E$ respectively. Prove $$ [PDBL] \cdot [PECM] \cdot [PFAN]=8\cdot [PFM] \cdot [PEL] \cdot [PDN] $$ Proposed by Steve Dinh