4.8. Find the angle at the vertex of an isosceles triangle if the medians drawn to the lateral sides are perpendicular to each other.
4.8. Let $A C=x$ (Fig. 4.7); since $\angle O A F=\angle A O F=45^{\circ}$, we have $O F=A F=\frac{x}{2}$. But $B O=2 O F=x$, hence $B F=\frac{3 x}{2}$. Assuming $\angle A B F=\beta$, we find $\operatorname{tg} \beta=\frac{A F}{B F}=\frac{\frac{x}{2}}{\frac{3 x}{2}}=\frac{1}{3}$.

Answer: $\operatorname{arctg} \frac{1}{3}$.