4.1. Through the vertex $A$ of the parallelogram $A B C D$, a line is drawn intersecting the diagonal $B D$, the side $C D$, and the line $B C$ at points $E, F$, and $G$ respectively. Find the ratio $B E: E D$, if $F G: F E=4$. Round your answer to the nearest hundredth if necessary.
Answer: 2.24 (exact value: $\sqrt{5}$).

Solution. Draw the line $E H$ parallel to $A B$ (point $H$ lies on side $B C$), and denote $B H=a, H C=b, B E=x, E D=y$ (see Fig. 1). By Thales' theorem, the desired ratio is $t=\frac{x}{y}=\frac{a}{b}$. From the similarity of triangles $B E G$ and $A E D$, we get $\frac{x}{a+5 b}=\frac{y}{a+b}$, so $t=\frac{x}{y}=\frac{a+5 b}{a+b}=\frac{t+5}{t+1} \Leftrightarrow t^{2}+t=t+5 \Leftrightarrow t^{2}=5$. Therefore, $t=\sqrt{5} \approx 2.24$.

![](https://cdn.mathpix.com/cropped/2024_05_06_72d2011c4e3388928cb1g-3.jpg?height=391&width=1062&top_left_y=90&top_left_x=500)

Fig. 1 (to problem 4)