8.7. Let's consider the equilateral triangle $A B C$. Points $D, E$, and $F$ are located on the sides $(B C),(C A)$, and, respectively, $(A B)$ such that the lines $D F$ and $A B$ are perpendicular, and the line $D E$ is the perpendicular bisector of the segment $C F$. Find the measure in degrees of the angle $D E F$.
Solution. Since points $D$ and $E$ are located on the perpendicular bisector of segment $(CF)$, it follows that $DC = DF$ and $EC = EF$ (see figure).

We obtain the congruence of triangles $EDC$ and $EDF$, from which we have the following equalities $m(\angle ACB) = m(\angle ECD) = m(\angle EFD) = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_366c775bb1927b312c81g-1.jpg?height=434&width=464&top_left_y=2079&top_left_x=1595)

Since triangle $BFD$ is right-angled at $F$, and triangle $CDF$ is isosceles with $DC = DF$, we then obtain the relations: $m(\angle ABC) = m(\angle FBD) = 60^{\circ}, m(\angle BDF) = 30^{\circ}$,

$$
m(\angle CDF) = 150^{\circ}, m(\angle DCF) = m(\angle DFC) = 15^{\circ}, m(\angle EDF) = 75^{\circ}
$$

Thus we obtain $\quad m(\angle DEF) = 180^{\circ} - m(\angle EFD) - m(\angle EDF) = 45^{\circ}$.