Task B-2.6. Let point $C$ be the vertex of the right angle in the right triangle $A B C$. On side $\overline{A C}$, a point $D$ is chosen such that $\angle A B D=\angle B A D=\alpha$. The length of side $\overline{A B}$ is $8 \mathrm{~cm}$, and the length of segment $\overline{A D}$ is $5 \mathrm{~cm}$. Determine the lengths of segments $\overline{B C}$ and $\overline{C D}$ and prove that for the angle $\alpha$ it holds that

$$
\sin (2 \alpha)=2 \sin \alpha \cos \alpha
$$
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9c6c142e457d4763ad17g-09.jpg?height=540&width=814&top_left_y=215&top_left_x=623)

Triangle $ABD$ is an isosceles triangle, and let $\overline{DE}$ be its altitude to the base $\overline{AB}$. The angle $\angle BDC$ is an exterior angle of triangle $ABD$ and its measure is $2\alpha$.

Image and/or explanation.

Then we have

$$
|BD|=|AD|=5 \text{ cm}, |ED|=3 \text{ cm}, |AE|=4 \text{ cm}
$$

From triangle $AED$ it follows that

$$
\sin \alpha = \frac{3}{5}, \cos \alpha = \frac{4}{5}
$$

Similarly, from triangle $ABC$ it follows that

$$
\sin \alpha = \frac{m}{8}, \cos \alpha = \frac{5+n}{8}
$$

From the last two equations, it follows that

$$
\frac{m}{8} = \frac{3}{5}, \frac{5+n}{8} = \frac{4}{5}
$$

Then we have

$$
m = \frac{24}{5} \text{ cm}, n = \frac{7}{5} \text{ cm}
$$

It follows that

$$
\sin (2\alpha) = \frac{m}{5} = \frac{24}{25} = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = 2 \sin \alpha \cos \alpha
$$