A triangle $A B C$ has side lengths $A B=50 \mathrm{~cm}, B C=20 \mathrm{~cm}$ and $A C=40 \mathrm{~cm}$. Let $M$ and $N$ be points on side $A B$ such that $C M$ is the angle bisector of $\angle A C B$ and $C N$ is the altitude to side $A B$. What is the measure, in centimeters, of $M N$?
Solution

By the Law of Cosines applied to the triangle $\triangle A B C$, we have

$$
\begin{aligned}
A C^{2} & =A B^{2}+B C^{2}-2 \cdot A B \cdot B C \cdot \cos \angle A B C \\
1600 & =2500+400-2 \cdot 50 \cdot 20 \cdot \cos \angle A B C \\
\cos \angle A B C & =\frac{13}{20}
\end{aligned}
$$

Moreover, by the Angle Bisector Theorem, we have

$$
\begin{aligned}
\frac{B M}{M A} & =\frac{B C}{A C} \\
\frac{B M}{B M+M A} & =\frac{B C}{B C+A C} \\
\frac{B M}{50} & =\frac{20}{60}
\end{aligned}
$$

Thus, $B M=50 / 3$. Finally,

$$
\begin{aligned}
M N & =B M-B N \\
& =\frac{50}{3}-B C \cdot \cos \angle A B C \\
& =\frac{50}{3}-20 \cdot \frac{13}{20} \\
& =\frac{11}{3}
\end{aligned}
$$