2. Let there be a triangle $A B C$, in which $|A B|=2|A C|$. On the sides $A B$ and $B C$ lie points $D$ and $E$ such that $\angle B A E=\angle A C D$. Segments $A E$ and $C D$ intersect at point $F$, and triangle $C F E$ is equilateral. Determine the sizes of the angles of triangle $A B C$.


I/2. 1. way. Since triangle $C E F$ is equilateral, $\angle E F C=60^{\circ}$. From this, it follows that $\angle C F A=$ $120^{\circ}$. Therefore, $\angle F A C=180^{\circ}-\angle C F A-\angle A C F=$ $60^{\circ}-\angle A C F$ and it holds that $\angle B A C=\angle B A E+\angle F A C=$ $\angle B A E+60^{\circ}-\angle A C D=60^{\circ}$.

Since $|A B|=2|A C|$ and $\angle B A C=60^{\circ}$, triangle $A B C$ is half of an equilateral triangle, so $\angle C B A=30^{\circ}$ and $\angle A C B=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-07.jpg?height=389&width=631&top_left_y=228&top_left_x=1204)

Conclusion $\angle C F A=120^{\circ}$

1 point

Calculation $\angle F A C=60^{\circ}-\angle A C F$

1 point

Conclusion $\angle B A C=60^{\circ}$

1 point

Observation that triangle $A B C$ is half of an equilateral triangle

2 points

Conclusion $\angle C B A=30^{\circ}$

1 point

Conclusion $\angle A C B=90^{\circ}$

1 point

2. way. Since triangle $C E F$ is equilateral, $\angle E F C=\angle C E F=60^{\circ}$. From this, it follows that $\angle C F A=$ $\angle A E B=120^{\circ}$. Since $\angle B A E=\angle A C D=$ $\angle A C F$, triangles $A B E$ and $C A F$ match in two angles, so they are similar. Therefore, $\frac{|A B|}{|A E|}=\frac{|A C|}{|C F|}$ or $|A E|=2|C F|$.

Triangle $C F E$ is equilateral, so $|A F|+$ $|F E|=|A E|=2|C F|=2|F E|$ or $|A F|=$

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-07.jpg?height=400&width=645&top_left_y=1048&top_left_x=1205)

$|F E|=|F C|$. Therefore, triangle $A F C$ is isosceles with vertex $F$ and since $\angle A F C=120^{\circ}$, $\angle F A C=\angle A C F=30^{\circ}$. From this, it follows that $\angle A C B=90^{\circ}, \angle B A C=60^{\circ}$ and $\angle C B A=30^{\circ}$.

Conclusion $\angle C F A=120^{\circ}$ 1 point

Observation that triangles $A B E$ and $C A F$ are similar. 1 point

Conclusion that $|A F|$ is equal to one of the lengths $|C E|,|E F|$ or $|C F|$ 1 point Triangle $A F C$ is isosceles 1 point Writing $\angle A C B=90^{\circ}, \angle B A C=60^{\circ}$ and $\angle C B A=30^{\circ}$