3. The isosceles trapezoid $A B C D$ has parallel sides $A B$ and $C D$. We know that $\overline{A B}=6, \overline{A D}=5$ and $\angle D A B=60^{\circ}$. A light ray is launched from $A$ that bounces off $C B$ at point $E$ and intersects $A D$ at point $F$. If $A F$ $=3$, calculate the area of triangle $A F E$.
Solution. Since the trapezoid is isosceles and $\angle D A B=60^{\circ}$, we can extend the sides $A D$ and $B C$ which intersect at $G$, thus forming an equilateral triangle $A B G$.

![](https://cdn.mathpix.com/cropped/2024_04_26_0518c9e7e0d4cca47497g-3.jpg?height=398&width=420&top_left_y=636&top_left_x=863)

Let $\alpha=\angle E A B$, then $\angle A E B=120-\alpha$. Since the ray exits symmetrically from the side $B C$, we have that $\angle G E F=120-\alpha$ and, therefore,

$$
\angle F E A=180-2(120-\alpha)=2 \alpha-60
$$

Since $\angle F A E=60-\alpha$, we have that $\angle A F E=180-\alpha$ and thus $\angle G F E=\alpha$. This proves that the triangles $G F E$ and $B A E$ are similar. Let $x=\overline{G E}$, then $\overline{E B}=6-x$. Since $\overline{F G}=\overline{A G}-\overline{A F}=6-3=3$, we have by the similarity of triangles that

$$
\frac{3}{x}=\frac{6}{6-x} \Longleftrightarrow 18-3 x=6 x \Longleftrightarrow 9 x=18
$$

from which we obtain that $x=2$ and, therefore, $\overline{G E}=2$ and $\overline{E B}=4$. We have

$$
A B / G F=6 / 3=2 \Longrightarrow \text { Area }(B A E) / \text { Area }(G F E)=2^{2}=4
$$

The area of the triangle $A F E$ can be calculated, for example,

$$
\begin{array}{r}
\text { Area }(A F E)=\text { Area }(A G B)-\text { Area }(G F E)-\text { Area }(A E B)= \\
\text { Area }(A G B)-5 \text { Area }(G F E)=\frac{1}{2} \cdot 6 \cdot 6 \cdot \sin 60-5 \cdot \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin 60= \\
(18-15) \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}
\end{array}
$$