4 [  In an isosceles triangle $A B C(A B=B C)$, the altitude $A F$ intersects the altitude $B D$ at point $O$, and $\frac{B O}{O D}=h \cdot$ In what ratio does the bisector $A E$ divide the altitude $B D$?
Let $O D=x, B O=h x$. Denote $\angle A B D=\angle C B D=\angle C A F=\alpha$. From the right triangles $A O D$ and $A B D$, we find that

$$
\operatorname{tg} \alpha=\frac{O D}{A D}=\frac{x}{A D}, \operatorname{tg} \alpha=\frac{A D}{B D}=\frac{A D}{x(h+1)}
$$

From the equality $\frac{x}{A D}=\frac{A D}{x(h+1)}$, it follows that $A D=x \sqrt{h+1}$. By the Pythagorean theorem,

$$
A B=\sqrt{A D^{2}+B D^{2}}=\sqrt{x^{2}(h+1)+x^{2}(h+1)^{2}}=x \sqrt{(h+1)(h+2)}
$$

Let $Q$ be the intersection point of the bisector $A E$ with the height $B D$. Then $A Q$ is the bisector of triangle $A B D$. By the property of the bisector of a triangle,

$$
\frac{B Q}{Q D}=\frac{A B}{A D}=\frac{x \sqrt{(h+1)(h+2)}}{x \sqrt{h+1}}=\sqrt{h+2}
$$

## Answer

$\sqrt{h+2}$, measured from vertex $B$.