The bisector BP of ∵AP perpendicular to ∠B is greater than P,

∠ headquarters base =∠EBP,

We also know that BP=BP, ∠ APB = ∠ BPE = 90,

In △ABP and △BEP,

∠ABP=∠EBP∠APB=∠BPE=90 BP=BP

∴△ABP≌△BEP(ASA)，

∴S△ABP=S△BEP,AP=PE，

∴△APC is as high as△ △CPE,

∴S△APC=S△PCE，

Let the area of δ δ△ACE be m,

∴S△ABE=S△ABC+S△ACE= 10+m

∴s△pbc= 12s△abe- 12s△ace= 10+m2-m2=5

Therefore, choose: B.