∵AD is the bisector of∝∠ ∝∠BAC,

∴∠EAD=∠OAD

∠∠OAD =∠ Addo

∴∠EAD=∠ADO

∴OD‖AE

∠∠AED = 90°

∴∠ODE=90

(2) connect OD. OC, OD=OA, ∠OAD =∠ official development assistance;

Let OG⊥AC cross AC to G, then AG=GC,

Yi Zheng △ OGA △ OGC (SSA)

∵DE⊥AC，

∴og‖de；

∵AD is the bisector of∝∠ ∝∠BAC,

∴∠BAC=2∠DAC=2∠OAD=2∠ODA，

∠∠BOD =∠OAD+∠ODA = 2∠OAD，

∴∠BOD=∠BAC，

∴od‖ae；

OD‖AE，ED‖OG

∴EDOG is a parallelogram.

∫∠DAC+∠ADE = 90 degrees,

∴∠ODA+∠ADE=90 degree

∴OGED is rectangular,

GE=OD=AB/2，

∠AEO =∠EOD∠AFE =∠OFD

∴△OFD∽△EFA

∴AE:DO=AF:DF

AF:DF =(AG+GE):(AB/2)=(AC/2+AB/2):(AB/2)= AC:AB+ 1 = 3:5+ 1 = 8:5