[ Angles between angle bisectors $\quad]$ 

In triangle $ABC$, the angle bisectors of the angles at vertices $A$ and $C$ intersect at point $D$. Find the radius of the circumcircle of triangle $ABC$, if the radius of the circumcircle of triangle $ADC$ with center at point $O$ is $R=6$, and $\angle ACO=30^{\circ}$.
Prove that $\angle A D C=90^{\circ}+\frac{1}{2} \angle B$.

## Solution

Since $\angle B A C+\angle A C B=180^{\circ}-\angle B$, then $\angle D A C+\angle A C D=90^{\circ}-\frac{1}{2} \angle B$. Therefore,

$$
\angle A D C=180^{\circ}-\left(90^{\circ}-\frac{1}{2} \angle B=90^{\circ}+\frac{1}{2} \angle B>90^{\circ}-\angle B,\right.
$$

i.e., angle $A D C$ is obtuse. Therefore, points $D$ and $O$ lie on opposite sides of the line $A C$. From the isosceles triangle $A O C$, we find that

$$
A C=2 O C \cos 30^{\circ}=2 R \cdot \frac{\sqrt{3}}{2}=6 \sqrt{3}
$$

Since $\angle A O C$ is a central angle of the circle with center $O$ and $\angle A O C=120^{\circ}$, the inscribed angle $\angle A D C$ is half the arc $A C$ not containing point $D$, i.e., $\angle A D C=\frac{1}{2} \cdot 240^{\circ}=120^{\circ}$. From the equality $\angle A D C=90^{\circ}+\frac{1}{2} \angle B$, we find that

$$
\angle B=2\left(\angle A D C-90^{\circ}\right)=2\left(120^{\circ}-90^{\circ}\right)=60^{\circ}
$$

Let $r$ be the radius of the circumcircle of triangle $A B C$. By the Law of Sines,

$$
r=\frac{A C}{2 \sin \angle B}=\frac{6 \sqrt{3}}{2 \cdot \frac{\sqrt{3}}{2}}=6
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-19.jpg?height=686&width=474&top_left_y=0&top_left_x=789)

## Answer

6.00
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