10.261. In an isosceles triangle \(ABC (AB = BC)\), a point \(D\) is taken on side \(BC\) such that \(BD: DC = 1: 4\). In what ratio does the line \(AD\) divide the height \(BE\) of triangle \(ABC\), counting from vertex \(B\)?
Solution.

Let $A D$ intersect $B E$ at point $F$ (Fig. 10.66). Draw $D K \| A C$. Since $\triangle B C E \sim \triangle B D K$, we have $\frac{B K}{B E}=\frac{K D}{E C}=\frac{B D}{B C}=\frac{1}{5}$, from which $B K=\frac{1}{5} B E, K D=\frac{1}{5} E C$. Since $\triangle K F D \sim \triangle A F E$, then $\frac{K D}{A E}=\frac{K F}{F E}=\frac{1}{5}$. Let $K F=x$; then $F E=5 x$ and $B K+x+5 x=B E$, from which $B K+6 x=5 B K$ or $6 x=4 B K$, i.e., $B K=\frac{3 x}{2}$.

Express $B F$ in terms of $x: B F=B K+x=\frac{3 x}{2}+x=\frac{5 x}{2}$. But $F E=5 x$ and, therefore, $B F: F E=1: 2$.

Answer: $1: 2$.