In $\triangle A B C$, given $A(5,-1), B(-1,7), C(1,2)$, find the length of the internal angle bisector $A D$ of $\angle A$ in $\triangle A B C$.
Solve $|A B|=\sqrt{(1+5)^{2}+(7+1)^{2}}=10,|A C|=\sqrt{(5-1)^{2}+(-1-2)^{2}}=5$.
By the property of the internal angle bisector, we have $\lambda=\frac{|B D|}{|\overline{C D}|}=\frac{|A B|}{|A \bar{C}|}=2$.
Let $D\left(x_{0}, y_{0}\right)$, by the coordinate formula of the section point, we know
$x_{0}=\frac{-1+2 \times 1}{1+2}=\frac{1}{3}, y_{0}=\frac{7+2 \times 2}{1+2}=\frac{11}{3}$. That is, $D\left(\frac{1}{3}, \frac{11}{3}\right)$,
Therefore,
$$
|A D|=\sqrt{\left(5-\frac{1}{3}\right)^{2}+\left(-1-\frac{11}{3}\right)^{2}}=\frac{14}{3} \sqrt{2}
$$