3. For the hexagon $A B C D E F$, it holds that $\angle B A F=150^{\circ}, \angle A C B=\angle A D C=90^{\circ}$, $|A C|=|B C|$, triangle $A B C$ is similar to triangle $A D E$ and triangle $B C D$ is similar to triangle $D E F$. Calculate the ratio of the lengths of segments $A B$ and $A F$.
Let $|A B|=a$. Since $A B C$ is an isosceles right triangle with vertex $C$, we have $|A C|=\frac{a}{\sqrt{2}}=\frac{a \sqrt{2}}{2}$ and $\angle A B C=\angle B A C=\frac{\pi}{4}$. Due to the similarity of triangles $A D E$ and $A B C$, it follows that $\angle A E D=\angle A C B=\frac{\pi}{2}$ and $|D E|=|A E|$.

From the similarity of triangles $B C D$ and $D E F$, we conclude $\angle B C D=\angle D E F$ and $\frac{|B C|}{|C D|}=\frac{|D E|}{|E F|}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-07.jpg?height=320&width=485&top_left_y=228&top_left_x=1365)

$$
\angle A C D=\angle B C D-\frac{\pi}{2}=\angle D E F-\frac{\pi}{2}=\angle A E F
$$

and

$$
\frac{|A C|}{|C D|}=\frac{|B C|}{|C D|}=\frac{|D E|}{|E F|}=\frac{|A E|}{|E F|}
$$

thus triangles $A C D$ and $A E F$ are similar, and $\angle C A D=\angle E A F$.

From

$$
150^{\circ}=\angle B A F=\angle B A C+\angle C A D+\angle C A E+\angle E A F=\frac{\pi}{2}+2 \angle C A D
$$

it follows that $\angle C A D=30^{\circ}$. Thus, triangle $C A D$ is half of an equilateral triangle, and $|A D|=\frac{|A C| \sqrt{3}}{2}=\frac{a \sqrt{6}}{4}$. Triangle $A D E$ is an isosceles right triangle, so $|A E|=\frac{|A D|}{\sqrt{2}}=\frac{a \sqrt{3}}{4}$. Considering that triangle $A E F$ is half of an equilateral triangle, we calculate $|A F|=\frac{|A E| \sqrt{3}}{2}=\frac{3 a}{8}$. We have $\frac{|A B|}{|A F|}=\frac{8}{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-07.jpg?height=77&width=1648&top_left_y=1286&top_left_x=204)

Writing the equality of angles $\angle B C D=\angle D E F$ or the ratio $\frac{|B C|}{|C D|}=\frac{|D E|}{|E F|} \ldots \ldots \ldots \ldots \ldots 1$ point

Conclusion: triangles $A C D$ and $A E F$ are similar .............................................

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-07.jpg?height=52&width=1650&top_left_y=1459&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-07.jpg?height=74&width=1648&top_left_y=1499&top_left_x=204)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-07.jpg?height=82&width=1648&top_left_y=1558&top_left_x=204)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-07.jpg?height=72&width=1650&top_left_y=1620&top_left_x=206)