3. In $\triangle A B C$, it is known that $D$ is a point on side $B C$, $\frac{B D}{D C}=\frac{1}{3}$, $E$ is the midpoint of side $A C$, $A D$ intersects $B E$ at point $O$, and $C O$ intersects $A B$ at point $F$. Find the ratio of the area of quadrilateral $B D O F$ to the area of $\triangle A B C$
For $\triangle A B C$ and point $O$, applying Ceva's Theorem, we get
$$
\frac{A F}{F B}=3, \frac{A F}{A B}=\frac{3}{4} \text {. }
$$

For $\triangle A D C$ and the transversal $F O C$, we get
$$
\begin{array}{l}
\frac{A O}{O D}=4, \frac{A O}{A D}=\frac{4}{5} . \\
\text { Hence } \frac{S_{\triangle A F O}}{S_{\triangle A B D}}=\frac{3}{5}, \frac{S_{\text {quadrilateral } B D O F}}{S_{\triangle \triangle B D}}=\frac{2}{5} . \\
\text { Also } \frac{S_{\triangle A B D}}{S_{\triangle A B C}}=\frac{1}{4}, \text { then } \frac{S_{\text {quadrilateral } B D D F}}{S_{\triangle A B C}}=\frac{1}{10} .
\end{array}
$$