Solution. Let $\angle A=\alpha, \angle B=\beta$. Point $O$ is the center of the inscribed circle of triangle $ABD$. $\angle BAO=\alpha / 4, \quad \angle ABO=\beta / 2$. 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, $\angle A=\angle C=72^{\circ}, \quad \angle B=36^{\circ}$. Let $AC=x, CD=y$. Triangles $ABC$ and $CAD$ are similar, $\frac{x}{y}=\frac{x+y}{x}, \quad x^{2}-xy-y^{2}=0$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-6.jpg?height=519&width=474&top_left_y=1705&top_left_x=1499)

By the condition $y=4, x=y(\sqrt{5}+1) / 2=2(\sqrt{5}+1)$.

Since $\cos (\angle ABC)=\cos (\angle DAC)=\cos \beta$, by the cosine rule for triangle $CAD$ we have $\cos 2\beta=\frac{y}{2x}=\frac{\sqrt{5}-1}{4}, \sin^{2} \beta=\frac{1}{2}\left(1-\frac{\sqrt{5}-1}{4}\right)=\frac{5-\sqrt{5}}{8}, \sin (\angle ABC)=\sqrt{\frac{5-\sqrt{5}}{8}}$. $S_{ABC}=\frac{1}{2}(x+y)^{2} \sin (\angle ABC)=\frac{(2\sqrt{5}+6)^{2}}{2} \sqrt{\frac{5-\sqrt{5}}{8}}=4(7+3\sqrt{5}) \sqrt{\frac{5-\sqrt{5}}{8}}=2 \sqrt{130+58\sqrt{5}}$.

Answer: $2 \sqrt{130+58\sqrt{5}}$.
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 the area of triangle $A B C$, if $C D=4$. The answer should not include trigonometric function notations and their inverses.