Another formula for the angle bisector in a triangle I have seen in an old geometry textbook that the formula for the length of the angle bisector at $A$ in $\triangle\mathit{ABC}$ is
\begin{equation*}
m_{a} = \sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]} ,
\end{equation*}
and I have seen in a much older geometry textbook that the formula for the length of the same angle bisector is
\begin{equation*}
m_{a} = \frac{2}{b + c} \sqrt{bcs(s - a)} .
\end{equation*}
($s$ denotes the semiperimeter of the triangle.)
I did not see such formulas in Euclid's Elements.  Was either formula discovered by the ancient Greeks?  May someone furnish a demonstration of either of them without using Stewart's Theorem and without using the Inscribed Angle Theorem?
 A: Here is a proof of the formula:
Let $AD$ be the angle bisector at $A$ (where $D \in BC$).
The area of $\triangle ABC$ is equal to the area of $\triangle ABD$ $+$ the area of $\triangle ACD$:
$$\frac{1}{2} bc \sin\alpha=\frac{1}{2} b m_a \sin(\alpha/2) + \frac{1}{2} c m_a \sin(\alpha/2)$$
$$m_a = \frac{2bc}{b+c}\cos(\alpha/2)$$
We can compute $\cos(\alpha/2)$ in terms of $a,b,c$ by using the cosine rule:
$$ 2\cos^2(\alpha/2) - 1=\cos\alpha=\frac{b^2 + c^2 - a^2}{2bc}$$
$$\cos^2(\alpha/2) = \frac{(b+c)^2 - a^2}{4bc} = \frac{s(s-a)}{bc} $$
Hence, we get: $$m_a = \sqrt{bc\left[ 1 - \left(\frac{a}{b+c}\right)^2\right]} = \frac{2}{b+c}\sqrt{bcs(s-a)}$$
A: We may prove it by avoiding the usual path, and proving other interesting things along the way.

Let $I$ be the incenter and $I_A$ the $A$-excenter. $\widehat{IBI_A}=\widehat{ICI_A}=\frac{\pi}{2}$, hence $IBI_A C$ is a cyclic quadrilateral. By Van Obel's theorem we have
$$ \frac{AI}{IL_A}=\frac{b+c}{a}, $$
hence it is enough to find $IA^2$. Let $C'$ be the symmetric of $C$ with respect to $AI$. We have:
$$ AI\cdot AI_A = \text{pow}_A\left(\Gamma_{I_A B C}\right)=AB\cdot AC'=bc.$$
The problem boils down to finding $II_A=AI_A-AI$, i.e. the diameter of the circumcircle of $I_A BC$. The midpoint of $II_A$ is also the midpoint of the $BC$-arc in the circumcircle of $BCI_A$, hence the previous diameter just depends on the length of $BC$, i.e. $a$, and the angle $\widehat{BNC}=\pi-\widehat{A}$. Putting everything together, we get that $IA$ is a root of
$$ x^2+x\frac{a}{\cos\frac{A}{2}}-bc, $$
but $IA\cos\frac{A}{2}=\frac{b+c-a}{2}$ and 
$$ \frac{a}{\cos^2\frac{A}{2}}=\frac{2a}{1+\cos A}=\frac{4abc}{(b+c-a)(b+c+a)},$$
hence:

$$ IA^2 = bc-\frac{2abc}{a+b+c} = \color{red}{bc-4rR}.$$

Now we may exploit the parallel axis theorem to compute the unpleasant squared distance $IG^2$ in a very slick way:
$$\begin{eqnarray*} \sum_{cyc}IA^2 = ab+ac+bc-12rR &=&3IG^2+GA^2+GB^2+GC^2\\ &=& 3IG^2+\frac{3}{4}(a^2+b^2+c^2)\end{eqnarray*}$$
leads to:

$$ IG^2 = ab+ac+bc-\frac{2abc}{a+b+c}-\frac{a^2+b^2+c^2}{4}$$

and we may prove $IO^2=R^2-2Rr$ in a similar way.
