Example 1 As shown in Figure 14-1, in quadrilateral $A B C D$, $\overrightarrow{A B}=\vec{a}-2 \vec{b}, \overrightarrow{C D}=3 \vec{a}-4 \vec{b}+$ $2 \vec{c}$, the midpoints of diagonals $A C$ and $B D$ are $E$ and $F$ respectively, find $\overrightarrow{E F}$.
Analysis From the expression, it can be seen that $\overrightarrow{A B}$ and $\overrightarrow{C D}$ are not collinear, so $\overrightarrow{E F}$ can definitely be represented by $\overrightarrow{A B}$ and $\overrightarrow{C D}$.
$$
\text { Solution } \begin{aligned}
\overrightarrow{E F} & =\overrightarrow{E A}+\overrightarrow{A B}+\overrightarrow{B F}=\frac{1}{2} \overrightarrow{C A}+\overrightarrow{A B}+\frac{1}{2} \overrightarrow{B D} \\
& =\frac{1}{2}(\overrightarrow{C D}+\overrightarrow{D A})+\overrightarrow{A B}+\frac{1}{2}(\overrightarrow{B A}+\overrightarrow{A D}) \\
& =\frac{1}{2} \overrightarrow{C D}+\frac{1}{2}(\overrightarrow{D A}+\overrightarrow{A D})+\overrightarrow{A B}+\frac{1}{2} \overrightarrow{B A} \\
& =\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{C D})=2 \vec{a}-3 \vec{b}+\vec{c} .
\end{aligned}
$$