3.4. The angle at the base of an acute isosceles triangle \(ABC (AB = BC)\) is \(\alpha\). In what ratio, counting from vertex \(A\), does the height \(BD\) divide the height \(AE\) ?
3.4. According to the condition, $\triangle ABC$ is acute-angled; hence, the point $K$ of intersection of the altitudes lies inside the triangle (Fig. 3.8). Let $AD = a$. From $\triangle AEC$ and $\triangle AKD$, we find $AE = 2a \sin \alpha$, $AK = \frac{a}{\sin \alpha} (\angle AKD = \angle C$, since both angles complement $\angle KAD$ to $90^{\circ}$). Further, we have

$$
\begin{gathered}
KE = AE - AK = 2a \sin \alpha - \frac{a}{\sin \alpha} = \\
\quad = \frac{a(2 \sin^2 \alpha - 1)}{\sin \alpha} = -\frac{a \cos 2\alpha}{\sin \alpha}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-502.jpg?height=529&width=479&top_left_y=105&top_left_x=676)

Fig. 3.8

Finally, we get

$$
\frac{AK}{KE} = -\frac{a}{\sin \alpha} : \frac{a \cos 2\alpha}{a \sin \alpha} = -\frac{1}{\cos 2\alpha}
$$