1. In an isosceles triangle \(ABC\), the base \(AC\) is equal to 1, and the angle \(ABC\) is equal to \(2 \operatorname{arctg} \frac{1}{2}\). Point \(D\) lies on side \(BC\) such that the area of triangle \(ABC\) is four times the area of triangle \(ADC\). Find the distance from point \(D\) to the line \(AB\) and the radius of the circumcircle of triangle \(ADC\).
Answer: distance $=\frac{3}{2 \sqrt{5}}, \quad$ radius $=\frac{\sqrt{265}}{32}$.