Angle bisector in triangle, quick question: $|AE| = \frac{bc}{a+c}$

Question

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?

Answer 1

Hint: We get $$\frac{AE}{EC}=\frac{c}{a}$$ And $$EC=b-AE$$ then we get $$\frac{b-AE}{AE}=\frac{c}{a}$$

Answer 2

let $\beta$ be half of the bisected angle, and $\theta$ denote the angle AEB.

then the sine rule in triangle AEB gives:

$$ \frac{|AE|}{\sin \beta} = \frac{c}{\sin \theta} $$

and the sine rule in triangle CEB gives:

$$ \frac{|EC|}{\sin \beta} = \frac{a}{\sin (\pi - \theta)} $$

since $\sin \theta = \sin (\pi-\theta)$ and $|AE|+|EC| = b$, the result follows

Answer 3

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