6. In triangle $A B C$, it is known that $A B=4, A C=6, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C$. ( $O$ - the center of the circumcircle of triangle $\left.A B C\right)$
Answer: $S_{O A_{2} C}=\frac{7}{\sqrt{3}} ; S_{A_{1} A_{2} C}=\frac{7 \sqrt{3}}{5}$.

Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+36-2 \cdot \frac{1}{2} \cdot 6 \cdot 4=28$, $B C=\sqrt{28}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C}{2 \sin 60^{\circ}}=\frac{\sqrt{28}}{\sqrt{3}}$. The angle $A_{2} O C$ is a central angle, so it is twice the angle $A_{2} A C$, which is given as $30^{\circ}$, hence $\angle A_{2} O C=60^{\circ}$. Therefore, the area of triangle $O A_{2} C$ is $\frac{\sqrt{3}}{4} \cdot R^{2}=\frac{7}{\sqrt{3}}$.

The bisector of a triangle divides the opposite side proportionally to the other two sides, so $A_{1} C: A_{1} B=A C: A B=6: 4$, from which $A_{1} C=\frac{3}{5} B C=\frac{3}{5} \sqrt{28}$. Triangle $O A_{2} C$ is equilateral, so $A_{2} C=R$. Angles $A_{2} C A_{1}$ and $A_{2} A B$ are equal as inscribed angles subtending the same arc, so $\angle A_{2} C A_{1}=30^{\circ}$. Thus, $S_{A_{1} A_{2} C}=\frac{1}{2} \cdot \frac{3}{5} \sqrt{28} \cdot \frac{\sqrt{28}}{\sqrt{3}} \cdot \sin 30^{\circ}=\frac{7 \sqrt{3}}{5}$.