3. As shown in Figure 1, $EF$ is the midline of $\triangle ABC$, $O$ is a point on $EF$, and satisfies $OE=2OF$. Then
the ratio of the area of $\triangle ABC$ to the area of $\triangle AOC$ is ( ).
(A) 2
(B) $\frac{3}{2}$
(C) $\frac{5}{3}$
(D) 3
3.D.

Since $EF$ is the midline of $\triangle ABC$, i.e., $EF \parallel AC$, therefore, $S_{\triangle COB}=S_{\triangle CPB}=\frac{1}{2} S_{\triangle ABC}$.
Thus, $S_{\triangle MOC}+S_{\triangle AOB}=\frac{1}{2} S_{\triangle ABC}$.
Also, $EO=2 FO$, so,
$\frac{S_{\triangle BOF}}{S_{\triangle COE}}=\frac{S_{\triangle MOF}}{S_{\triangle AOE}}=2$, which means $\frac{S_{\triangle AOC}}{S_{\triangle AOB}}=2$.
Therefore, $S_{\triangle ABC}=3 S_{\triangle AOC}$.