(1)S 1=19a;
(2)过点C作CG⊥BE于点G,
设S △BPF=x,S △APE=y,
∵ S △BPC =
1
2 BP•CG=70 ; S △PCE =
1
2 PE•CG=35 ,
∴
S △BPC
S △PCE =
1
2 BP•CG
1
2 PE•CG =
70
35 =2 .
∴
BP
EP =2 ,即BP=2EP.
同理,
S △APB
S △APE =
BP
PE .
∴S △APB=2S △APF.
∴x+84=2y.①
∵
S △APB
S △BPD =
AP
PD =
x+84
40 ,
S △APC
S △PCD =
AP
PD =
y+35
30 ,
∴
x+84
40 =
y+35
30 .②
由①②,得
x=56
y=70 ,
∴S △ABC=315.
(3)设S △BPF=m,S △APE=n,如图所示.
依题意,得S △APF=S △APC=m,S △BPC=S △BPF=m.
∴S △PCE=m-n.
∵
S △APB
S △APE =
S △BPC
S △PCE =
BP
PE ,
∴
2m
n =
m
m-n .
∴2m(m-n)=mn,
∵m≠0,
∴2m-2n=n.
∴
n
m =
2
3 .
∴
S △APE
S △BPF =
2
3 .