Problem 4. Given is $\triangle A B C$ with

$$
\measuredangle B A C=40^{\circ}, \measuredangle A B C=20^{\circ} \text { and } \overline{A B}-\overline{B C}=5 \text{ cm}
$$

If the angle bisector of $\measuredangle A C B$ intersects $A B$ at point $M$, find the length of $C M$.
Solution. Let $N \in AB$ such that $\overline{BC}=\overline{BN}$. Then $\triangle BCN$ is isosceles, so $\measuredangle BNC=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ}$. From $CM$ being the angle bisector of $\measuredangle ACB=120^{\circ}$, it follows that $\measuredangle MCB=60^{\circ}$, i.e.

$$
\measuredangle BMC=180^{\circ}-20^{\circ}-60^{\circ}=100^{\circ}
$$

from which

$$
\measuredangle NMC=180^{\circ}-100^{\circ}=80^{\circ}
$$

Thus, $\triangle NCM$ is isosceles, so

![](https://cdn.mathpix.com/cropped/2024_06_05_e235c78af50361de0ec1g-08.jpg?height=276&width=606&top_left_y=1304&top_left_x=877)

$\measuredangle NCM=180^{\circ}-80^{\circ}-80^{\circ}=20^{\circ}$.

Furthermore, from

$$
\measuredangle ACN=\measuredangle ACM-\measuredangle NCM=60^{\circ}-20^{\circ}=40^{\circ}
$$

it follows that $\triangle ANC$ is isosceles, i.e., $\overline{AN}=\overline{NC}$. Then

$$
\overline{CM}=\overline{NC}=\overline{AN}=5 \mathrm{~cm}
$$