On the sides of triangle $ABC$, points $D, E, F, G$ are given, as shown in the figure. It is given that quadrilateral $DEFG$ is a rhombus and segments $AD, DE$, and $EB$ are congruent.

Determine the measure of angle $ACB$.

(I. Jančigová)

![](https://cdn.mathpix.com/cropped/2024_04_17_d278c4c03a37be47319bg-2.jpg?height=409&width=876&top_left_y=1323&top_left_x=590)

Hint. What is the relative position of lines $AC$ and $DF$?
According to the assumptions, segments $A D$ and $G F$ are parallel and congruent. Therefore, quadrilateral $A D F G$ is a parallelogram, and in particular, lines $A G$ and $D F$ are parallel.

By similar reasoning, it can be shown that lines $B F$ and $E G$ are also parallel.

Since the diagonals $D F$ and $E G$ of rhombus $D E F G$ are perpendicular, lines $A G$ and $B F$ are also perpendicular. Point $C$ is the intersection of these lines, so angle $A C B$ is a right angle.

![](https://cdn.mathpix.com/cropped/2024_04_17_d278c4c03a37be47319bg-2.jpg?height=394&width=870&top_left_y=2167&top_left_x=593)

Another solution. According to the assumptions, lines $A B$ and $G F$ are parallel, so angles $D A G$ and $F G C$ are congruent (corresponding angles). Furthermore, triangle $A D G$ is isosceles ($A D$ and $D G$ are congruent), so angles $D A G$ and $D G A$ are congruent. We denote these three congruent angles by the symbol $\alpha$, see the figure.

Similarly, it can be shown that angles $E B C, G F C$, and $E F B$ are congruent; we denote these angles by $\beta$.

From the parallelism of lines $A B$ and $G F$, it also follows that angles $A D G$ and $D G F$ are congruent (alternate interior angles). Furthermore, opposite angles in parallelogram $D E F G$ are congruent, in particular, $D G F$ is congruent with $D E F$. We denote these three congruent angles by the symbol $\delta$.

Similarly, angles $B E F, E F G$, and $E D G$ are congruent; we denote these angles by $\varepsilon$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d278c4c03a37be47319bg-3.jpg?height=394&width=874&top_left_y=745&top_left_x=591)

Among the angles mentioned, several relationships hold, from which we select:

- $\varepsilon=2 \alpha$ (exterior angle of triangle $A D G$),
- $\varepsilon+2 \beta=180^{\circ}$ (sum of interior angles of triangle $E B F$),
- $\alpha+\beta+\gamma=180^{\circ}$ (sum of interior angles of triangle $G F C$).

From the first two equalities, it follows that $2 \alpha+2 \beta=180^{\circ}$, so $\alpha+\beta=90^{\circ}$. Substituting into the third equality, we get $\gamma=180^{\circ}-\alpha-\beta=90^{\circ}$. Therefore, angle $A C B$ is a right angle.