1. In $\triangle A B C$, $D$ is a point on $B C$, $\frac{B D}{D C}=\frac{1}{3}$, $E$ is the midpoint of $A C$. $A D$ intersects $B E$ at $O$, and $C O$ intersects $A B$ at $F$. Find the ratio of the area of quadrilateral $B D O F$ to the area of $\triangle A B C$.
1. For $\triangle A B C$ and point $O$, by Ceva's Theorem, we get $\frac{A F}{F B}=3, \frac{A F}{A B}=\frac{3}{4}$. Also, for $\triangle A D C$ and the transversal $F O C$, by Menelaus' Theorem, we get $\frac{A O}{O D}=4, \frac{A O}{A D}=\frac{4}{5}$, hence $\frac{S_{\triangle A F O}}{S_{\triangle A B D}}=\frac{3}{4} \cdot \frac{4}{5}=\frac{3}{5}$, from which we know $\frac{S_{B D O F}}{S_{\triangle A B D}}=\frac{2}{5}$. Also, $\frac{S_{\triangle A B D}}{S_{\triangle A B C}}=\frac{1}{4}$, so $\frac{S_{B D O F}}{S_{\triangle A B C}}=\frac{2}{5} \cdot \frac{1}{4}=$ $\frac{1}{10}$.