In trapezoid $ABCD$, $AD\parallel BC$ and $AD:BC=1:2$. Point $F$ lies on $AB$ and point $E$ is on $CF$. If $S_{\triangle{AOF}}:S_{\triangle{DOE}}=1:3$ and $S_{\triangle{BEF}}=24$, find the area of $\triangle{AOF}$.
Firstly, we have
$$S_{\triangle{ADE}}-S_{\triangle{ADF}}=S_{\triangle{ODE}}-S_{\triangle{OAF}}=2S_{\triangle{OAF}}$$
and $$S_{\triangle{ADE}}-S_{\triangle{ADF}} = \frac{1}{2}\cdot h\cdot AD$$
Therefore, we have $$S_{\triangle{AOF}}=\frac{1}{4}\cdot h\cdot AD$$
Meanwhile, we have
$$S_{\triangle{BEF}}=S_{\triangle{BCF}}-S_{\triangle{BCE}}=\frac{1}{2}\cdot h\cdot BC$$
Consequently, we have
$$S_{\triangle{AOF}}=S_{\triangle{BEF}}\cdot \frac{1}{2}\cdot \frac{AD}{BC}=S_{\triangle{BEF}}\cdot \frac{1}{2}\cdot \frac{1}{2}=\boxed{4}$$