填:6 S△PDE=_6_S△PDG∵AD平分∠BAC∴∠BAD=∠CAD∵PE∥AB,PF∥AC∴∠BAD=∠EPD ,∠CAD=∠FPD∴∠EPD=∠FPD∴点D在∠EPF的平分线上∴D到PE的距=D到PF的距离相∵PE:PF=3:2∴S△PDE:S△PDF=3:2∴S△PDE=3/5 *S△PEF∵点G是EF的中点∴S△PGE=1/2 *S△PEF∴S△PDG=S△PDE-S△PGE=3/5 *S△PEF -1/2 *S△PEF=1/10 *△PEF∴S△PDE/S△PDG=(3/5)/(1/10)=6∴S△PDE=6S△PDGO(∩_∩)O~