3. The hypotenuse $A B$ 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 $E$ be the intersection of side $A B$ and line $C D$. Calculate the length of segment $A E$.
$\mathbf{I} / 3$. Triangle $A B C$ is half of an equilateral triangle, so in it, $|B C|=\frac{1}{2}|A B|$. Let $\Varangle A C D = \varphi$. Then $\Varangle D C B = \Varangle A C B - \Varangle A C D = \frac{\pi}{2} - \varphi$ and therefore

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

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-07.jpg?height=280&width=491&top_left_y=223&top_left_x=1359)

Since $\Varangle D B A = \Varangle A C D = \varphi$, we have $60^{\circ} = \Varangle C B A = \Varangle D B A + \Varangle C B D = 2 \varphi$ and from this $\varphi = 30^{\circ}$. Since $\Varangle C B E = 60^{\circ}$ and $\Varangle E C B = \Varangle D C B = \frac{\pi}{2} - \varphi = 60^{\circ}$, triangle $E B C$ is equilateral. It follows that $|E B| = |B C| = \frac{1}{2}|A B|$ and $|A E| = |A B| - |E B| = \frac{1}{2}|A B| = 5 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-07.jpg?height=69&width=1636&top_left_y=685&top_left_x=216)

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

Conclusion that triangle $E B C$ is equilateral ............................................ 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-07.jpg?height=63&width=1633&top_left_y=839&top_left_x=220)