<s>[INST] 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.)
 [/INST] 
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
</s>