How to construct a triangle given two sides and their bisector? Suppose I have two triangle sides $AB$ and $AC$, and the length of the angle bisector of $A$. How can I construct (straightedge and compass) the triangle? (This question is from one of the earlier Moscow Math Olympiads.)
 A: By Stewart's theorem the length $\ell$ of the angle bisector through $A$ fulfills
$$ \ell^2 =\frac{bc}{(b+c)^2}\left((b+c)^2-a^2\right)$$
hence
$$a\sqrt{bc}= (b+c)\sqrt{bc-\ell^2}$$
and it is enough to construct three segments with lengths $\sqrt{bc},(b+c),\sqrt{bc-\ell^2}$ to find $a$ (then $ABC$) through the intersecting chords theorem:

A: 
Let
the length 
of the angle bisector of $A$
$|AD|=d$
and the side lengths 
$|BC|=a,|AC|=b,|AB|=c$,
$|BD|=m$.
Then according to
Stewart's Theorem
for $\triangle ABC$,
\begin{align} 
c^2(a-m)+b^2m
-a(d^2+(a-m)m)
&=0
\tag{1}\label{1}
.
\end{align}  
By the law of sines,
\begin{align} 
\triangle ABD:\quad
\frac{m}{\sin\tfrac\alpha2}
&=
\frac{c}{\sin\delta}
,\\
\triangle ADC:\quad
\frac{a-m}{\sin\tfrac\alpha2}
&=
\frac{b}{\sin\delta}
,\\
\text{hence, }\quad
\frac{m}{a-m}&=\frac{c}{b}
,\\
a&=\frac{m(c+b)}{c}
,
\end{align}  
and \eqref{1} becomes
\begin{align} 
\frac{m(c+b)(c^2b-m^2b-d^2c)}{c^2}
&=0
,\\
\end{align}
\begin{align} 
m^2&=
\frac{c(bc-d^2)}{b}
.
\end{align}
Given $m$ we can found that
\begin{align} 
a^2&=\frac{(bc-d^2)(b+c)^2}{bc}
,
\end{align}
hence, the length of the missing side $a$
can be constructed from known values $b,c,d$,
for example, using 
Intersecting_chords_theorem
A: We can find the length of side BC as follows.
Let $AD = m$ be the angle bisector such that $D$ is on $BC$ and $\angle BAB = \angle CAD = \theta$ .  Since $AD$ is angle bisector of $\angle A $ therefore $${AB \over AC} = {BD \over DC} \implies \frac cb = {BD\over DC} $$ Let $BD = kc$ and $DC = kb$. Now using cosine rule in $\Delta CAD$ and $\Delta BAD $ we get 
$$(kb)^2 = b^2 + m^2 - 2bm \cos \theta $$ and
$$ (kc)^2 = c^2 + m^2 - 2cm \cos \theta $$ Now equating $\cos\theta$ in both equation, we get
$$ {k^2b^2-b^2-m^2 \over -2bm} = {k^2c^2-c^2-m^2 \over -2cm}$$ Simplyfying the above equation we will get $$ k = \sqrt{{bc - m^2 \over bc}} $$ So $$BC = \sqrt{{bc-m^2 \over bc}}(c+b)$$ Now we have all sides of $\Delta ABC$.
