Solution. a) The fact that point $D$ lies on the circle is irrelevant. The areas of triangles $A B O$ and $O B C$ are in the ratio $A O: O C$. The same applies to triangles $A O D$ and $D O C$. Therefore,

$$
S_{A B D}: S_{B C D}=A O: O C=A B: B C=5: 8
$$

by the property of the angle bisector.

![](https://cdn.mathpix.com/cropped/2024_05_06_9993983e5654471c2fedg-2.jpg?height=503&width=512&top_left_y=2144&top_left_x=1503)
b) 1st method. Let the angle $A B D$ be $\alpha, A B=a=5, B C=b=8, A C=c$. Clearly, $D$ is the midpoint of the arc $A C$, so $A D=C D=\frac{A C}{2} / \cos \alpha=\frac{c}{2 \cos \alpha}$. By Ptolemy's theorem, $B D \cdot c=a \cdot C D+b \cdot A D=(a+b) \frac{c}{2 \cos \alpha}$, from which $B D=\frac{a+b}{2 \cos \alpha}$

We have

$$
S_{A B D}=\frac{1}{2} a \cdot B D \sin \alpha=\frac{a(a+b)}{4} \operatorname{tg} \alpha
$$

This value can take any positive value for a suitable $\alpha$.

2nd method. From the previously proven condition, $S_{A B D}=S_{0}$ is equivalent to $S=S_{A B C D}=\frac{a+b}{a} S_{0}$.

Let's calculate the area of the quadrilateral $A B C D$.

$$
\begin{gathered}
S=\frac{1}{2} a b \sin 2 \alpha+\frac{1}{2} A D \cdot C D \sin 2 \alpha=\frac{1}{2}\left(a b+\frac{1}{4} \frac{c^{2}}{\cos ^{2} \alpha}\right) \sin 2 \alpha= \\
=\frac{4 a b \cos ^{2} \alpha+c^{2}}{8 \cos ^{2} \alpha} \sin 2 \alpha=\left(4 a b \cos ^{2} \alpha+a^{2}+b^{2}-2 a b \cos 2 \alpha\right) \frac{\sin 2 \alpha}{8 \cos ^{2} \alpha}= \\
\left(a^{2}+b^{2}+2 a b\left(2 \cos ^{2} \alpha-\cos 2 \alpha\right)\right)=\frac{(a+b)^{2}}{4} \operatorname{tg} \alpha
\end{gathered}
$$

This value can take all possible positive values for suitable $\alpha$.

Grading criteria: A complete solution to part a) with the correct answer 16 - 10 points. The correct answer in part b) without rigorous or incomplete proofs - 0 points. A complete solution to part b) with the condition(s) specified when it is possible. For example, stating that $\operatorname{tg} \alpha=80 / 13$ or $B D=\sqrt{6569} / 2-10$ points.
Problem 4. Given a triangle $ABC$ with sides $AB=5$; $BC=8$. The bisector of angle $B$ intersects the circumcircle of the triangle at point $D$. a) It is known that the area of triangle $ABD$ is 10. Find the area of triangle $BCD$. b) Can it be that $S_{ABD}$ equals 100?