<s>[INST] 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?
 [/INST] 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: $$m_a = \sqrt{bc\left[ 1 - \left(\frac{a}{b+c}\right)^2\right]} = \frac{2}{b+c}\sqrt{bcs(s-a)}$$
</s>