Draw a parallel line of alternating current, pass through point F, and intersect with BE at point G.

1∴⊿BFG∽⊿BAE。

∴FG:AE=BG:BE=BF:BA=2:3.

AE = EC .

∴FG:CE=2:3.

2∵⊿FGP∽⊿CEP and FG: CE = 2: 3.

∴ FP: CP = GP = EP = 2: 3。 ∴ CP: CF = 3: 5。 That is CP = (3/5) cf.

Then BG: ge = BF: fa = 2: 1 (parallel lines are proportional)

And gp: EP = 2 ∶ 3 means BP: be = 12: 15 = 4/5.

∴BP=(4/5)BE.

∴ To sum up, as vectors, there are: BP=(4/5)BE, and CP = (3/5) CF.

3 (all the following are vectors)

According to the triangle rule of vector addition, BE+EA=BA.

∴be=ba-ea=(-ab)-(-ae)=-ab+ae=-a+( 1/2)b.

That is, the vector BE=-a+( 1/2)b and the vector BP=(4/5)BE.

∴ Vector BP = (-4/5) A+(2/5) B.

According to the triangle rule of vector addition, CF+FA=CA.

∴ vector cf = ca-fa = (-AC)-(-af) =-b+af =-b+(1/3) ab = (1/3) a-b.

That is, the vector CF=( 1/3)a-b and the vector CP = (3/5) cf.

∴ vector CP = (1/5) A-(3/5) B