如图,梯形ABCD中,AD∥BC,两腰BA与CD的延长线相交于P,PF⊥BC,AD=3.6,BC=6,EF=3,则PF=_.

问题描述:

如图,梯形ABCD中,AD∥BC,两腰BA与CD的延长线相交于P,PF⊥BC,AD=3.6,BC=6,EF=3,则PF=______.

∵AD∥BC,PF⊥BC,
∴PE⊥AD,
∵AD∥BC,
∴△PAD∽△PBC,

AD
BC
=
PE
PF

3.6
6
=
PF−3
PF

∴PF=7.5.
故答案为:7.5.