In triangle $A B C, \angle A=50^{\circ}, \angle B=60^{\circ}$, and $\angle C=70^{\circ}$. A ray of light bounces from point $D$ on $B C$ to $E$ on $C A$ to $F$ on $A B$ and then back to $D$. Find the angles of $\triangle D E F$.

(Light always takes the shortest path between points, meaning it bounces off an edge at equal angles.)
Let $\angle B D F=\angle C D E=x, \angle C E D=\angle A E F=y$, and $\angle A F E=$ $\angle B F D=z$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7e55c3ead5cd8924307ag-2.jpg?height=716&width=1011&top_left_y=333&top_left_x=634)

Since $\triangle A E F, \triangle B D F$, and $\triangle C D E$ all have angle sum $180^{\circ}$, we get

$$
\begin{aligned}
& x+y+70^{\circ}=180^{\circ}, \\
& x+z+60^{\circ}=180^{\circ}, \\
& y+z+50^{\circ}=180^{\circ} .
\end{aligned}
$$

Adding the three equations, subtraction $180^{\circ}$ from both sides, and dividing by 2 gives

$$
x+y+z=180^{\circ} \text {. }
$$

From this, we get

$$
x=50^{\circ}, y=60^{\circ}, z=70^{\circ} .
$$

(Thus, the triangles $\triangle A E F, \triangle B D F$, and $\triangle C D E$ are in fact all congruent to $\triangle A B C$ !) Now we get

$$
\begin{aligned}
& \angle E D F=180^{\circ}-2 \cdot 50^{\circ}=80^{\circ}, \\
& \angle D E F=180^{\circ}-2 \cdot 60^{\circ}=60^{\circ}, \\
& \angle D F E=180^{\circ}-2 \cdot 70^{\circ}=40^{\circ} .
\end{aligned}
$$