4. In triangle $A B C$, it is known that $A B=3, A C=4, \angle B A C=60^{\circ}$. 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 $A_{1} A_{2} C \cdot(O$ - the center of the circumcircle of triangle $A B C)$.
Answer: $S_{\triangle O A_{2} C}=\frac{13}{4 \sqrt{3}} ; S_{\triangle A_{1} A_{2} C}=\frac{13}{7 \sqrt{3}}$.

Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+9-2 \cdot \frac{1}{2} \cdot 3 \cdot 4=13$, $B C=\sqrt{13}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C}{2 \sin 60^{\circ}}=\frac{\sqrt{13}}{\sqrt{3}}$. 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 $\frac{\sqrt{3}}{4} \cdot R^{2}=\frac{13}{4 \sqrt{3}}$.

The bisector of a triangle divides the opposite side proportionally to the other two sides, so $A_{1} C: A_{1} B=A C: A B=4: 3$, from which $A_{1} C=\frac{4}{7} B C=\frac{4}{7} \sqrt{13}$. 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 $\angle A_{2} C A_{1}=30^{\circ}$. Therefore, $S_{A_{1} A_{2} C}=\frac{1}{2} \cdot \frac{4}{7} \sqrt{13} \cdot \frac{\sqrt{13}}{\sqrt{3}} \cdot \sin 30^{\circ}=\frac{13}{7 \sqrt{3}}$.