## Task B-4.5.

A piece of paper has the shape of an equilateral triangle $ABC$ with side length $15 \, \text{cm}$. We fold the paper so that vertex $A$ comes to point $D$ on side $\overline{BC}$, with $|BD| = 3 \, \text{cm}$. This creates a fold $\overline{EF}$, where point $E$ is on $\overline{AB}$ and point $F$ is on $\overline{AC}$. Determine the length of the fold $|\overline{EF}|$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-20.jpg?height=679&width=766&top_left_y=303&top_left_x=568)

Precise image with marked congruent segments $x, y$:

Apply the Law of Cosines to triangles $C D F, B E D$, and $D E F$.

From $\triangle C D F$:

$$
\begin{aligned}
& y^{2}=(15-y)^{2}+12^{2}-2 \cdot 12 \cdot(15-y) \cdot \cos 60^{\circ} \\
& y^{2}=225-30 y+y^{2}+144-180+12 y \\
& 18 y=189 \\
& y=\frac{21}{2}
\end{aligned}
$$

From $\triangle B E D$:

$$
\begin{aligned}
& x^{2}=(15-x)^{2}+3^{2}-2 \cdot(15-x) \cdot 3 \cdot \cos 60^{\circ} \\
& x=7
\end{aligned}
$$

From $\triangle D E F$:

$$
\begin{aligned}
& z^{2}=x^{2}+y^{2}-2 x y \cos 60^{\circ} \\
& z^{2}=7^{2}+\left(\frac{21}{2}\right)^{2}-2 \cdot 7 \cdot \frac{21}{2} \cdot \frac{1}{2} \\
& z=\frac{7 \sqrt{7}}{2}
\end{aligned}
$$

Therefore, the length of the fold $|\overline{E F}|$ is $\frac{7 \sqrt{7}}{2} \mathrm{~cm}$.