4. On the side $B C$ of the isosceles triangle $A B C(\overline{A B}=\overline{B C})$, a point $D$ is chosen such that $\overline{B D}: \overline{D C}=1: 4$. In what ratio does the line $A D$ divide the height $B E$ from vertex $B$?
Solution. Let the intersection of $B E$ and $A D$ be denoted by $F$. Let $K \in B E$ be such that $D K \| C A$. According to this, $\triangle B C E \sim \triangle B D K$ and

$$
\frac{\overline{B K}}{\overline{B E}}=\frac{\overline{K D}}{\overline{C E}}=\frac{\overline{B D}}{\overline{B C}}=\frac{1}{5}
$$

that is,

$$
\overline{B K}=\frac{1}{5} \overline{B E} \text { and } \overline{K D}=\frac{1}{5} \overline{C E}
$$

Since triangles $D K F$ and $A E F$ have equal angles, they are similar, so according to this

$$
\frac{\overline{K D}}{\overline{A E}}=\frac{\overline{\overline{K F}}}{\overline{F E}}=\frac{1}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_cc33540ece97b25c9e0cg-2.jpg?height=366&width=394&top_left_y=1473&top_left_x=1072)

If we introduce the notation $\overline{K F}=x$, then $\overline{F E}=5 x$, so from the equation

$$
\overline{B K}+\overline{K F}+\overline{F E}=\overline{B E}
$$

we get $\overline{B K}=\frac{3}{2} x$. Since

$$
\overline{B F}=\overline{B K}+\overline{K F}=\frac{3}{2} x+x=\frac{5}{2} x
$$

from the equation $\overline{F E}=5 x$ we get $\overline{B F}: \overline{F E}=\frac{5}{2} x: 5 x=1: 2$.

## II year