<s>[INST] The Length of a Bisector How can I prove that : 
$$b_{A}=\frac{1}{b+c} \cdot \sqrt{bc\left[(b+c)^2-a^2\right]} ?$$ where $b_{A}$ is the length of the bisector from $A$ . 
Thanks :) 
 [/INST] Let $P$ be the point where the bisector of $\angle A$ meets $BC$. Let $x=BP$ and $y=PC$. Then $x+y=a$. Moreover, by a standard theorem on angle bisectors, we have 
$\dfrac{x}{c}=\dfrac{y}{b}$, or equivalently $bx=cy$. 
We have two linear equations in two unknowns. Solve for $x$ and $y$. We obtain
$$x=\frac{ac}{b+c}\qquad\text{and}\qquad y=\frac{ab}{b+c}.$$
Now let $\theta=\angle APB$ and $\phi=\angle APC$. Then $\cos\phi=-\cos\theta$. By the Cosine Law on $\triangle BPA$, we have
$$c^2=b_A^2+x^2 -2b_A \,x\cos\theta.\tag{$1$}$$
Similarly 
$$b^2=b_A^2 +y^2+2b_A  \,y\cos\theta.\tag{$2$}$$
Multiply both sides of $(1)$ by $y$, and both sides of $(2)$ by $x$, and add. We get
$$yc^2+xb^2=(x+y)b_A^2+xy(x+y).$$
Substituting our values for $x$ and $y$, and noting that $x+y=a$, we obtain
$$\frac{abc^2}{b+c}+\frac{acb^2}{b+c}=ab_A^2+\frac{a^3bc}{(b+c)^2}.$$
This simplifies to
$$bc=b_A^2+\frac{a^2bc}{(b+c)^2},$$
or equivalently
$$b_A^2=\frac{1}{(b+c)^2}\left(bc\left[(b+c)^2-a^2\right]\right),$$
which is what we wanted to show.
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