3. In triangle $A B C$, we have

$$
|\angle B A C|=40^{\circ},|\angle C B A|=20^{\circ} \text { and }|A B|-|B C|=10
$$

The angle bisector of $\angle A C B$ intersects the segment $\overline{A B}$ at point $M$. Determine the length of the segment $\overline{C M}$.
## First solution.

Let $E$ be a point on the segment $\overline{A B}$ such that $|B E|=|B C|$.

We have $|A B|-|B C|=|A B|-|B E|=|A E|$, so $|A E|=10 \mathrm{~cm}$.

Two angles in triangle $A B C$ are known, so the third angle is $|\angle A C B|=120^{\circ}$.

Since $C M$ is the angle bisector of that angle, we have $|\angle A C M|=|\angle M C B|=60^{\circ}$.

Due to the choice of point $E$, triangle $E B C$ is isosceles, so $|\angle B E C|=|\angle E C B|=80^{\circ}$.

Further,

$|\angle E C M|=|\angle E C B|-|\angle M C B|=20^{\circ}$.

$|\angle C M E|=80^{\circ}$

Since $|\angle C M E|=|\angle M E C|$, triangle $E M C$ is isosceles, so $|\mathrm{EC}|=|\mathrm{MC}|$.

Also, $|\angle A C E|=|\angle A C M|-$

![](https://cdn.mathpix.com/cropped/2024_05_30_38ebdf2cda2a1fe1ae2eg-4.jpg?height=377&width=1049&top_left_y=131&top_left_x=945)
$|\angle E C M|=40^{\circ}$, that is, $|\angle A C E|=|\angle E A C|$. Therefore, triangle AEC is isosceles, so $|\mathrm{AE}|=|\mathrm{EC}|$.

Connecting the proven equalities, we get $|M C|=|E C|=|A E|=10 \mathrm{~cm}$.