The question is:
Prove that the square of the length, '$d$' of the angle bisector $AL$
of $\angle A$ meeting $BC$ at $L$ in a $\Delta ABC$ is $$bc \left(1- \left (\frac{a}{b+c}\right)^2\right).$$
I found this problem in a worksheet on the topic Stewart's theorem. But any other ways of solving may also be used. I'm sorry, I could not provide a diagram for this question as the site said it had some problems with that.
What I have tried:
I labelled $BL$ as $m$ and $LC$ as $n$. So according to Stewart's theorem,
$$a(mn+d^2)=b^2m+c^2n$$
I rearranged it to get $d^2$ on the LHS, as that is what we need. So I got
$$d^2= \frac{b^2m+c^2n-amn}{a}. $$
$$ d^2=b^2\cdot\frac{m}{a}+c^2\cdot\frac{n}{a}-mn$$
Then taking $m=a-n$, I got
$$d^2=b^2-\frac{n}{a}(b^2-c^2)-an+n^2.$$
I am not able to get rid of $n$. I tried using the cosine rule but that didn't help.
Any thoughts over this?
Best Answer
By the Stewart's theorem
\begin{align} c^2u+b^2(a-u)&=a(d^2+(a-u)u) ,\\ d^2&=\frac 1a\,(c^2u+b^2(a-u))-(a-u)u \tag{1}\label{1} , \end{align}
by the angle bisector theorem,
\begin{align} b\,(a-u)&=c\,u \tag{2}\label{2} ,\\ u&=\frac {ab}{b+c} \tag{3}\label{3} ,\\ a-u&= \frac{ac}{b+c} \tag{4}\label{4} . \end{align}
Combination of \eqref{1} with \eqref{3} and \eqref{4} gives \begin{align} d^2& =\frac{bc((b+c)^2-a^2)}{(b+c)^2} =bc\left(1-\frac{a^2}{(b+c)^2}\right) . \end{align}