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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