Prof. Aliyev is back, this time with a geometry problem. As shown in Figure 4, $ABC$ is a triangle. Points $F$ and $E$ lie on side AB, with $F$ closer to $B$. Point $D$ lies on side $AC$. Point $G$ is the intersection of segments $BD$ and $CE$, and $H$ is the intersection of ray $FG$ with side $AC$. Prove it must always be the case in this situation that

$$\frac{BF}{FE}>\frac{DH}{HC},$$

where in this inequality, BF (for example) denotes the length of the corresponding segment.