Problem 3. If the lengths of two sides of a triangle are $6 \mathrm{~cm}$ and $12 \mathrm{~cm}$, and the angle between them is $120^{\circ}$, what is the length of the angle bisector of that angle?
Solution. Let $AM$ be the bisector of $\measuredangle BAC = 120^{\circ}$ (see diagram). Then

![](https://cdn.mathpix.com/cropped/2024_06_05_d4fd26fa5c16caf575adg-8.jpg?height=356&width=914&top_left_y=716&top_left_x=401)

$\measuredangle BAM = \measuredangle MAC = 60^{\circ}$. Let $N$ be a point on the extension of side $AB$ such that $\overline{AN} = \overline{AC} = 6 \text{ cm}$. Since $\measuredangle NAC = 180^{\circ} - 120^{\circ} = 60^{\circ}$, it follows that triangle $NAC$ is equilateral. Then from $\measuredangle CNA = \measuredangle MAB = 60^{\circ}$, we get that $CN$ and $MA$ are parallel segments. By applying Thales' theorem for proportional segments, we get that $\overline{AM} : \overline{NC} = \overline{AB} : \overline{NB}$, from which it follows that

$$
\overline{AM} = \frac{\overline{AB} \cdot \overline{NC}}{\overline{NB}} = \frac{12 \cdot 6}{6 + 12} = 4 \text{ cm}
$$