## Task B-4.6.

In triangle $A B C$, the lengths of the sides are $|\overline{A B}|=7,|\overline{B C}|=8$ and $|\overline{A C}|=9$. Point $D$ is on side $\overline{A C}$ such that $\varangle C B D=45^{\circ}$. Determine the length of segment $\overline{B D}$ and the ratio of the areas of triangles $A B C$ and $D B C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8e0ddf1d7bf4bfb115dcg-18.jpg?height=526&width=609&top_left_y=568&top_left_x=655)

Let $\gamma=\angle BCA, \delta=\angle BDC, x=|\overline{BD}|$.

By the cosine rule, we have

$$
\begin{gathered}
\cos \gamma=\frac{8^{2}+9^{2}-7^{2}}{2 \cdot 8 \cdot 9}=\frac{2}{3} \\
\sin \gamma=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}
\end{gathered}
$$

By the sine rule, we have

$$
\begin{array}{cc}
\frac{x}{\sin \gamma}=\frac{8}{\sin \delta}, & 1 \text { point } \\
x=\frac{8 \sin \gamma}{\sin \delta}=\frac{8 \sin \gamma}{\sin \left(180^{\circ}-\left(45^{\circ}+\gamma\right)\right)}=\frac{8 \sin \gamma}{\sin \left(45^{\circ}+\gamma\right)}=\frac{8 \sin \gamma}{\sin 45^{\circ} \cos \gamma+\cos 45^{\circ} \sin \gamma}, & 2 \text { points } \\
x=\frac{8 \cdot \frac{\sqrt{5}}{3}}{\frac{\sqrt{2}}{2} \cdot \frac{2}{3}+\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{5}}{3}}=\frac{8 \sqrt{10}}{2+\sqrt{5}}=8 \sqrt{10}(\sqrt{5}-2) . & 1 \text { point }
\end{array}
$$

The ratio of the areas is

$$
\begin{aligned}
\frac{P(ABC)}{P(DBC)} & =\frac{8 \cdot 9 \sin \gamma}{x \cdot 8 \sin 45^{\circ}}=\frac{9 \cdot \frac{\sqrt{5}}{3}}{8 \sqrt{10}(\sqrt{5}-2) \frac{\sqrt{2}}{2}} \\
& =\frac{3}{8(\sqrt{5}-2)}=\frac{3(\sqrt{5}+2)}{8}
\end{aligned}
$$