## Task B-3.5.

The lengths of the sides of triangle $A B C$ are $|B C|=4 \mathrm{~cm}$ and $|A C|=5 \mathrm{~cm}$, and the length of the part of the angle bisector of $\varangle A C B$ that lies within the triangle is $s=\frac{10}{3} \mathrm{~cm}$. Calculate the length of side $\overline{A B}$.
## Solution.

The bisector of an angle in a triangle intersects the opposite side in the ratio of the remaining sides (The Angle Bisector Theorem of a Triangle), i.e., $\frac{x}{y}=\frac{5}{4}$, where $x=|A D|, y=|D B|$, and point $D$ is the intersection of the given bisector and side $\overline{A B}$. Then, $x=5 k, y=4 k$.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-21.jpg?height=731&width=962&top_left_y=500&top_left_x=453)

Apply the Law of Cosines to triangles $A D C$ and $C D B$.

$$
\cos \frac{\gamma}{2}=\frac{5^{2}+\left(\frac{10}{3}\right)^{2}-25 k^{2}}{2 \cdot 5 \cdot \frac{10}{3}}, \quad \cos \frac{\gamma}{2}=\frac{\left(\frac{10}{3}\right)^{2}+4^{2}-16 k^{2}}{2 \cdot 4 \cdot \frac{10}{3}}
$$

By equating the right-hand sides of these equations, we get

$$
\frac{5^{2}+\left(\frac{10}{3}\right)^{2}-25 k^{2}}{2 \cdot 5 \cdot \frac{10}{3}}=\frac{\left(\frac{10}{3}\right)^{2}+4^{2}-16 k^{2}}{2 \cdot 4 \cdot \frac{10}{3}}
$$

which simplifies to $k^{2}=\frac{4}{9}$,

so $k=\frac{2}{3}$.

Then, $|A B|=x+y=9 k=6 \text{ cm}$.

Note: A similar solution can be obtained by applying the Law of Cosines to triangles $A D C$ and $A B C$ to determine $\cos \alpha$.

If a student does not use the Angle Bisector Theorem of a Triangle, they can arrive at the ratio $\frac{x}{y}=\frac{5}{4}$ by applying the Law of Sines to triangles $A D C$ and $D B C$, which is worth 2 points:

$$
\frac{x}{\sin \frac{\gamma}{2}}=\frac{5}{\sin \varphi}, \quad \frac{y}{\sin \frac{\gamma}{2}}=\frac{4}{\sin \left(180^{\circ}-\varphi\right)}, \quad \text{where } \varphi=\angle A D C
$$

Since $\sin \left(180^{\circ}-\varphi\right)=\sin \varphi$, it follows that $\frac{x}{y}=\frac{5}{4}$, and we proceed as in the first solution.