23. In $\triangle A B C, A B: A C=4: 3$ and $M$ is the midpoint of $B C$. $E$ is a point on $A B$ and $F$ is a point on $A C$ such that $A E: A F=2: 1$. It is also given that $E F$ and $A M$ intersect at $G$ with $G F=72 \mathrm{~cm}$ and $G E=x \mathrm{~cm}$. Find the value of $x$.
23. Ans: 108
Let the area of $\triangle A G F$ be $S_{1}$ and the area of $\triangle A G E$ be $S_{2}$. Let the area of $\triangle A M C$ be $S$, so that the area of $\triangle A B M$ is also $S$, since $M$ is the midpoint of $B C$. Now,
$$
\frac{S_{1}}{S}=\frac{A G \cdot A F}{A M \cdot A C} \text { and } \frac{S_{2}}{S}=\frac{A G \cdot A E}{A M \cdot A B} .
$$

Therefore,
$$
\frac{S_{2}}{S_{1}}=\frac{A G \cdot A E \cdot A M \cdot A C}{A M \cdot A B \cdot A G \cdot A F}=\frac{3}{4} \times 2=\frac{3}{2} .
$$

Note that $\frac{S_{2}}{S_{1}}=\frac{G E}{G F}=\frac{x}{72}$. Therefore, $\frac{3}{2}=\frac{x}{72} \Longrightarrow x=108$.