如图,已知平行四边形ABCD,对角线AC和BD相交于点O,点P在边AD上,过点P作PE垂直于AC,

点P是AD的中点,点F是DO的中点
则PF为三角形AOD的中线,PF∥AC
PF⊥BD,AC⊥BD
所以□ABCD为正方形
设BC=x
BD=√(BC^2+CD^2)=√2x
BO=OD=1/2BD=√2/2x
OF=1/2OD=1/2*√2/2x=√2/4x
BF=BO+OF=√2/2x+√2/4x=3√2/4x
BF-BC=3-2√2,3√2/4x-x=3-2√2
x=4(3-2√2)/(3√2-4)=4(3-2√2)(3√2+4)/[(3√2-4)(3√2+4)]
=4(9√2+12-12-8√2)/[18-16]
=2√2
BC=2√2