4. In triangle $A B C$, the bisector $A D$ is drawn. It is known that the centers of the inscribed circle of triangle $A B D$ and the circumscribed circle of triangle $A B C$ coincide. Find $C D$, if $A C=\sqrt{5}+1$. The answer should not include trigonometric function notations or their inverses.

(20 points)
Solution: Let $\angle A=\alpha, \quad B=\beta$. Point $O$ is the center of the inscribed circle of triangle $ABD$. $\angle BAO=\alpha / 4, \angle ABO=\beta / 2$. Since $O$ is the center of the circumscribed circle around triangle $ABC$, triangle $AOB$ is isosceles, and $\angle BAO=\angle ABO, \beta=\alpha / 2$. Triangles $AOC$ and $BOC$ are isosceles, and $\angle ACO=3 \alpha / 4, \angle BCO=\alpha / 4$. Since $\angle A+\angle B+\angle C=180^{\circ}$, then $\alpha+\alpha / 2+3 \alpha / 4+\alpha / 4=180^{\circ}$, and $\alpha=72^{\circ}$. Triangle $ABC$ is isosceles, $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$.

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$\frac{x}{y}=\frac{x+y}{x}, \quad x^{2}-x y-y^{2}=0$.

By the condition $x=\sqrt{5}+1, y=x(\sqrt{5}-1) / 2, \quad y=2$.

Answer: 2.

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