Angle bisector in triangle, quick question: $|AE| = \frac{bc}{a+c}$ Triangle $ABC$; $AB=c, BC=a, AC=b$; angle bisector of angle $(c, a)$ cuts $AC$ in point $E$.
Why is the following true?
$$|AE| = \frac{bc}{a+c}$$
Where does that come from?

The angle bisector theorem:
$$\frac{CE}{AE}=\frac ac$$
Then:
$$\frac{bc}{a+c}=\frac{b}{\frac ac+1}=\frac{b}{\frac{CE}{AE}+1}=\frac{AE\cdot b}{CE+AE}=AE$$
