Two congruent equilateral triangles $A B C$ and $B D E$ are given such that the measure of angle $A B D$ is greater than $120^{\circ}$ and less than $180^{\circ}$, and points $C, E$ lie in the same half-plane defined by the line $A D$. The intersection of $C D$ and $A E$ is denoted as $F$.

Determine the measure of angle $A F D$.

(I. Jančigová)

![](https://cdn.mathpix.com/cropped/2024_04_17_e988acd5633232ee9496g-1.jpg?height=457&width=803&top_left_y=2076&top_left_x=632)

Hint. You know the sums of the interior angles in triangles or polygons.
The internal angle at vertex $A$, resp. $D$ in quadrilateral $A F D B$ is also an internal angle of triangle $A B E$, resp. $D B C$. Triangles $A B E$ and $D B C$ are isosceles and congruent by assumption. Therefore, the internal angles at vertices $A, E, D, C$ in these triangles are congruent. The size of these angles depends on the angle $A B D$, which is not precisely defined. Let's denote the angles in question as shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_04_17_e988acd5633232ee9496g-2.jpg?height=463&width=805&top_left_y=682&top_left_x=631)

The sum of the internal angles in triangle $A B E$, resp. $D B C$ gives the relationship between $\alpha$ and $\omega$:

$$
2 \alpha + \omega - 60^{\circ} = 180^{\circ}.
$$

The sum of the internal angles in quadrilateral $A F D B$ gives the relationship for the sought angle $\varphi$:

$$
2 \alpha + \omega + \varphi = 360^{\circ}.
$$

From the first relationship, we get $2 \alpha + \omega = 240^{\circ}$, which together with the second relationship gives $\varphi = 120^{\circ}$.

The size of angle $A F D$ is $120^{\circ}$.

Another solution. The internal angle in an equilateral triangle has a size of $60^{\circ}$. When rotating around point $B$ by this angle in the clockwise direction, point $A$ is mapped to $C$ and point $E$ is mapped to $D$. Thus, line $A E$ is mapped to line $C D$.

Therefore, the angle $A F C$, resp. $E F D$ has a size of $60^{\circ}$. The sought angle $A F D$ complements this angle to a straight angle, so it has a size of $120^{\circ}$.

Note. The result $\varphi = 120^{\circ}$ indeed does not depend on the size of angle $\omega$. The restrictions in the problem are not essential and serve only to make the problem easier to grasp. In limiting cases, we get special and usually simpler cases: for $\omega = 120^{\circ}$, points $C, E$, and $F$ coincide and $\varphi$ is the sum of two internal angles of equilateral triangles, for $\omega = 180^{\circ}$ we get a problem similar to problem $\mathbf{Z 7}$-I-2.