$解：∵四边形ABCD是平行四边形$
$∴AB=CD$
$易得△DEF∽△CEB，△DEF∽△ABF$
$∴\frac {S_{△DEF}}{S_{△CEB}}={(\frac {DE}{CE})}^{2}，\frac {S_{△DEF}}{S_{△ABF}}={(\frac {DE}{AB})}^{2}$
$∵CD=2DE$
$∴DE∶CE=1∶3，DE∶AB=1∶2$
$∵S_{△DEF}=a$
$∴S_{△CEB}=9a，S_{△ABF}=4a$
$∴S_{四边形BCDF}=S_{△CEB}-S_{△DEF}=8a$
$∴S_{▱ABCD}=S_{四边形BCDF}+S_{△ABF}=8a+4a=12a$
