## Task B-2.5.

In parallelogram $A B C D$, the angle at vertex $B$ is $120^{\circ}$ and side $A D$ is three times the length of side $A B$. A point $E$ is chosen on side $B C$, and a point $F$ is chosen on side $A D$ such that $A B E F$ is a rhombus. What is the cosine of angle $E A C$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-17.jpg?height=294&width=1059&top_left_y=1052&top_left_x=424)

Let $|A B|=x, \angle E A C=\varphi$. By applying the cosine rule to triangle $A B E$, we get:

$|A E|^{2}=|A B|^{2}+|B E|^{2}-2 \cdot|A B| \cdot|B E| \cos 120^{\circ}=x^{2}+x^{2}-2 \cdot x \cdot x \cdot \cos 120^{\circ}=3 x^{2}$, 1 point

so $|A E|=x \sqrt{3}$.

By applying the cosine rule to triangle $A B C$, we get:

$|A C|^{2}=|A B|^{2}+|B C|^{2}-2 \cdot|A B| \cdot|B C| \cos 120^{\circ}=x^{2}+(3 x)^{2}-2 \cdot x \cdot 3 x \cdot \cos 120^{\circ}=13 x^{2}$,

so $|A C|=x \sqrt{13}$.

By applying the cosine rule to triangle $AEC$, we get:

$$
\cos \varphi=\frac{|A C|^{2}+|A E|^{2}-|E C|^{2}}{2 \cdot|A C| \cdot|A E|}=\frac{13 x^{2}+3 x^{2}-4 x^{2}}{2 \cdot x \sqrt{13} \cdot x \sqrt{3}}=\frac{2 \sqrt{39}}{13}
$$

Note: The conclusion that $|A E|=x \sqrt{3}$ could also be reached from the fact that triangle $A B E$ is isosceles and the relation $\sin 60^{\circ}=\frac{\frac{1}{2}|A E|}{x}$, which also earns 2 points.