5. In $\triangle A B C$, let $D$ and $E$ be the trisection points of $B C$, with $D$ between $B$ and $E$, $F$ be the midpoint of $A C$, and $G$ be the midpoint of $A B$. Let $H$ be the intersection of line segments $E G$ and $D F$. Find the ratio $E H$ : $H G$.
5. Let $\overrightarrow{G H}=a \overrightarrow{G E}, \overrightarrow{F H}=b \overrightarrow{F D}$, then $\overrightarrow{A H}=\overrightarrow{A G}+a \overrightarrow{G E}=\overrightarrow{A F}+b \overrightarrow{F D}$. Also, $\overrightarrow{A G}=\frac{1}{2} \overrightarrow{A B}$, so $\overrightarrow{G E}=\overrightarrow{G B}+\overrightarrow{B E}=\frac{1}{2} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{B C}=\frac{1}{2} \overrightarrow{A B}+\frac{2}{3}(\overrightarrow{A C}-\overrightarrow{A B})=-\frac{1}{6} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{A C}$. From $\overrightarrow{A F}=$ $\frac{1}{2} \overrightarrow{A C}$, we get $\overrightarrow{F D}=\frac{2}{3} \overrightarrow{A B}-\frac{1}{6} \overrightarrow{A C}$, thus $\frac{1}{2} \overrightarrow{A B}+a\left(-\frac{1}{6} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{A C}\right)=\frac{1}{2} \overrightarrow{A C}+$ $b\left(\frac{2}{3} \overrightarrow{A B}-\frac{1}{6} \overrightarrow{A C}\right)$, which simplifies to $\left(\frac{1}{2}-\frac{a}{6}-\frac{2 b}{3}\right) \overrightarrow{A B}-\left(\frac{1}{2}-\frac{b}{6}-\frac{2 a}{3}\right) \overrightarrow{A C}=\overrightarrow{0}$. Since $\overrightarrow{A B}$ and $\overrightarrow{A C}$ are not collinear, we have $\left\{\begin{array}{l}\frac{1}{2}-\frac{a}{6}-\frac{2 b}{3}=0 \\ \frac{1}{2}-\frac{b}{6}-\frac{2 a}{3}=0,\end{array}\right.$ solving which gives $a=b=\frac{3}{5}$, hence $E H: H G=2: 3$.