3. The cathetus $A C$ of the right triangle $A B C$ with a right angle at $C$ is 1 dm long, and the angle $\angle BAC$ is $30^{\circ}$. Inside the triangle, there is a point $D$ such that $\angle B D C=90^{\circ}$ and $\angle A C D=\angle D B A$. Let $F$ be the intersection of side $A C$ and line $B D$. Calculate the length of segment $A F$.
II/3. Triangle $A B C$ is half of an equilateral triangle, so in it, $|A C|=|A B| \frac{\sqrt{3}}{2}$, which gives us $|A B|=\frac{2}{\sqrt{3}}|A C|$ and $|B C|=\frac{1}{2}|A B|=\frac{\sqrt{3}}{3}|A C|$.

Let $\Varangle A C D=\Varangle D B A=\varphi$. Then $\Varangle D C B=$ $\Varangle A C B-\Varangle A C D=\frac{\pi}{2}-\varphi$ and therefore

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-09.jpg?height=274&width=488&top_left_y=226&top_left_x=1361)

$$
\Varangle D B C=\pi-\Varangle B D C-\Varangle D C B=\pi-\frac{\pi}{2}-\left(\frac{\pi}{2}-\varphi\right)=\varphi .
$$

Thus, $60^{\circ}=\Varangle C B A=\Varangle C B D+\Varangle D B A=2 \varphi$, so $\varphi=30^{\circ}$.

Since $\Varangle F B C=\Varangle D B C=30^{\circ}$, triangle $B F C$ is similar to triangle $A B C$. Therefore, $|C F|:|B C|=|B C|:|A C|$, from which it follows that

$$
|C F|=\frac{|B C|^{2}}{|A C|}=\frac{\frac{1}{3}|A C|^{2}}{|A C|}=\frac{1}{3}|A C|
$$

Thus, $|A F|=|A C|-|C F|=\frac{2}{3}|A C|=\frac{2}{3} \mathrm{dm}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-09.jpg?height=72&width=1627&top_left_y=1049&top_left_x=220)

Calculation that angles $\Varangle C B D$ and $\Varangle D B A$ are equal and thus both $30^{\circ} \ldots \ldots \ldots \ldots \ldots . .2$ points

Mentioned similarity of triangles $B F C$ and $A B C \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-09.jpg?height=62&width=1633&top_left_y=1214&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-09.jpg?height=63&width=1633&top_left_y=1262&top_left_x=220)