$$d^2=\frac{bc}{(b+c)^2}\left[(b+c)^2-a^2\right]\tag{1}$$
where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.
Lemma 1 (Tehebycheff). Let $\triangle{ABC}$ be a triangle. Let $AD$ be the angle bisector of $\angle{BAC}$ in $\triangle{ABC}$. Let $d$ be the length of $AD$ and $\angle{BAD}=\angle{CAD}=\frac{\alpha}{2}$. Then $d$ is given by:
$$d=\frac{2bc\cos{\frac{\alpha}{2}}}{b+c}\tag{2}$$
Proof. Let $BD=m$ and $CD=n$. We make use of the Angle Bisector Theorem and the Law of Cosines. Indeed, by the Angle Bisector Theorem we have
$$\frac{b}{c}=\frac{n}{m}$$
which can also be written as
$$\frac{b^2}{c^2}=\frac{n^2}{m^2}\tag{3}$$
Now, by the Law of Cosines,
$$n^2=b^2+d^2-2bd\cos{\frac{\alpha}{2}}\tag{4}$$
$$m^2=c^2+d^2-2cd\cos{\frac{\alpha}{2}}\tag{5}$$
Dividing $(4)$ by $(5)$ and substituting $\frac{n^2}{m^2}$ by $\frac{b^2}{c^2}$ we have
$$\frac{b^2}{c^2}=\frac{b^2+d^2-2bd\cos{\frac{\alpha}{2}}}{c^2+d^2-2cd\cos{\frac{\alpha}{2}}}$$
From which we get
$$d^2(b^2-c^2)=2bcd\cos{\frac{\alpha}{2}}(b-c)$$
Isolating $d$, factoring and simplifying we obtain $(2)$.
$\square$
Another proof of Lemma 1 can be found here.
Lemma 2. Let $\triangle{ABC}$ be a triangle and let $a$, $b$ and $c$ be the lengths of sides $BC$, $AC$ and $BC$, respectively. Then the following identity holds
$$2\cos{\frac{\alpha}{2}}=\sqrt{\frac{(b+c)^2-a^2}{bc}}\tag{6}$$
A proof of Lemma 2 can be found here.
$\square$
Main proof
Combining $(2)$ and $(6)$ we obtain
$$d=\frac{bc}{(b+c)}\cdot{\sqrt{\frac{(b+c)^2-a^2}{bc}}}$$
Raising both sides to the power of 2 we get $(1)$.
$\square$
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