设AC与BD交于点O。可得S△AOD=3，OA=OD=5/2.不管动点P在AD上何处，总有S△AOP＋S△DOP=S△AOD1/2(OA·PF＋OD·PE)=(1/2)×OA×(PE＋PF)=3故PE＋PF=12/5
