32. Express the length of a) the median and b)* the bisector of a triangle in terms of the lengths of its sides.
32. a) Let $m_{a}$ denote the length of the median [ $A M$ ] of triangle $A B C$. By drawing lines through vertices $B$ and $C$ parallel to the opposite sides of the triangle, we obtain a parallelogram with sides of lengths $b$ and $c$, and diagonals of lengths $2 m_{a}$ and $a$, where $a, b, c$ are the lengths of the sides of the triangle. Applying the formula derived in problem $29, b$, we get:

$$
\left(4 m_{a}^{2}+a^{2}=2 b^{2}+2 c^{2}\right) \Rightarrow\left(m_{a}^{2}=\frac{b^{2}+c^{2}}{2}-\frac{a^{2}}{4}\right)
$$

b) According to the formula from the solution of problem 5 (§ 2), we have:

$$
\vec{n}=\frac{\vec{b}+\overrightarrow{c b}}{b+c}
$$

Squaring both sides of the equation scalarly, we get:

$$
n^{2}=\frac{2 b^{2} c^{2}+2 b c \cdot \overrightarrow{b c}}{(b+c)^{2}}
$$

But from triangle $A B C$ we have

$$
\left(a^{2}=b^{2}+c^{2}-2 \vec{b} \vec{c}\right) \Rightarrow\left(2 \vec{b} \vec{c}=b^{2}+c^{2}-a^{2}\right)
$$

Substituting this value into formula (1), we get:

$$
n^{2}=\frac{c b^{2} c^{2}+b c\left(b^{2}+c^{2}-a^{2}\right)}{(b+c)^{2}}=b c \frac{(b+c)^{2}-a^{2}}{(b+c)^{2}}
$$

Similarly, for the length $n$ of the external angle bisector, we get:

$$
n^{2}=b c \frac{a^{2}-(c-b)^{2}}{(c-b)^{2}}
$$