Two congruent equilateral triangles $A B C$ and $B D E$ are given such that points $A, B$, $D$ lie on the same line and points $C, E$ lie in the same half-plane defined by that line. The intersection of $C D$ and $A E$ is marked as $F$.

Determine the measure of angle $A F D$.

(I. Jančigová)

![](https://cdn.mathpix.com/cropped/2024_04_17_8fdde125de0a7cae0b19g-1.jpg?height=434&width=808&top_left_y=1782&top_left_x=630)

Hint. It is useful to focus on groups of congruent angles.
The internal angle at vertex $A$, resp. $D$ in triangle $A F D$ 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 $A$ and $D$ are congruent.

Thus, triangle $A F D$ is also isosceles, and the angle $A F D$ is congruent to the angle $A B E$, resp. $D B C$. This angle is an external angle of an equilateral triangle, so it has a measure of $120^{\circ}$.

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

![](https://cdn.mathpix.com/cropped/2024_04_17_8fdde125de0a7cae0b19g-2.jpg?height=443&width=826&top_left_y=618&top_left_x=615)

Another solution. The internal angle in an equilateral triangle has a measure 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, the line $A E$ is mapped to the line $C D$.

The angle $A F C$, resp. $E F D$ therefore has a measure of $60^{\circ}$. The desired angle $A F D$ complements this angle to a right angle, so it has a measure of $120^{\circ}$.

Notes. From the assumptions, a number of other relationships can be derived, which may lead to alternative justifications for the previous result. We briefly mention a few observations (for additional notation, see the figure below):

The quadrilateral $A B E C$, resp. $B D E C$ is a rhombus. The diagonals in a rhombus bisect the corresponding internal angles and are perpendicular to each other.

- From the first fact, it follows that the angle $B A E$, resp. $B D C$ has a measure of $30^{\circ}$. The angle $A F D$, which is the remaining internal angle in triangle $A F D$, therefore has a measure of $120^{\circ}$ (the sum of the internal angles of a triangle is $180^{\circ}$).
- From the second fact, we know the internal angles at vertices $G$ and $H$ in quadrilateral $B H F G$. The internal angle at vertex $B$ has a measure of $60^{\circ}$. The angle $A F D$, which is the remaining internal angle in quadrilateral $B H F G$, therefore has a measure of $120^{\circ}$ (the sum of the internal angles of a quadrilateral is $360^{\circ}$).

The given figure is part of an equilateral triangle $A D I$, where points $B, C, E$ are the midpoints of its sides. Segments $A E, C D$, and $I B$ are altitudes and also axes of symmetry of this triangle. Triangles $A F D, D F I$, and $I F A$ are congruent and together form triangle $A D I$. Their internal angles at vertex $F$ therefore have a measure of $120^{\circ}$ (one-third of a full angle).

![](https://cdn.mathpix.com/cropped/2024_04_17_8fdde125de0a7cae0b19g-3.jpg?height=762&width=826&top_left_y=287&top_left_x=615)