解：∵$\frac {AE}{EC}=\frac {1}{2}$
∴$\frac {AE}{AC}=\frac {1}{3}$
∵$DE∥BC$
∴$∠DEO=∠OBC，$$∠EDO=∠OCB$
∴$△DOE∽△COB$
∴$\frac {DE}{BC}=\frac {AE}{AC}=\frac {1}{3}$
∴$\frac {{C}_{△DOE}}{{C}_{△BOC}}=\frac {DE}{BC}=\frac {1}{3}，$$\frac {{S}_{△DOE}}{{S}_{△BOC}}={(\frac {DE}{BC})}^2={(\frac {1}{3})}^2=\frac {1}{9}.$
