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?

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.
