Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?

Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it
 A: Let $|AB|=c$, $|BC|=a=c+2$,
$|AC|=b=5$, 
$\angle BCA=\gamma$,
$\angle CAB=\alpha=2\gamma$
By the sine rule we have
\begin{align} 
\frac{\sin\alpha}{a}
&=
\frac{\sin\beta}{b}
=\frac{\sin\gamma}{c}
,\\
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin(\pi-3\gamma)}{5}
=\frac{\sin\gamma}{c}
.
\end{align}  
By the rules based on 
componendo and dividendo,
\begin{align} 
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin\gamma}{c}
=\frac{\sin2\gamma-\sin\gamma}{c+2-c}
=\frac{\sin2\gamma-\sin\gamma}{2}
,\\
\frac{\sin(3\gamma)}{5}
&=
\frac{\sin2\gamma-\sin\gamma}{2}
,
\end{align}
\begin{align} 
2\sin(3\gamma)
-
5\sin2\gamma+5\sin\gamma
&=0
,\\
8\sin\gamma\cos^2\gamma-2\sin\gamma
-10\sin\gamma\cos\gamma+5\sin\gamma
&=0
,\\
8\cos^2\gamma
-10\cos\gamma+3
&=0
,
\end{align}  
so $\cos\gamma$ 
must be either $\tfrac12$ or $\tfrac34$.
It follows that possible values for $\gamma$
are $60^\circ$ or $\arccos\tfrac34\approx41.41^\circ$. 
But since $\beta=180^\circ-3\gamma$, 
$\gamma=60^\circ$ results in $\beta=0$, a degenerate case,
so the only suitable choice is
\begin{align} 
\cos\gamma&=\tfrac34
,\\
\frac{\sin\gamma}c&=
\frac{\sin2\gamma-\sin\gamma}2
,\\
c&=\frac{2\sin\gamma}{\sin2\gamma-\sin\gamma}
,\\
&=
\frac{2\sin\gamma}{2\sin\gamma\cos\gamma-\sin\gamma}
=
\frac{2}{2\cos\gamma-1}
=
\frac{2}{2\cdot\tfrac34-1}
=4.
\end{align}
Thus $\triangle ABC$ has sides $4,5$ and $6$cm.
A: Hint: with the Theorem of cosines we get
$$c^2=(c+2)^2+25-2(c+2)\cdot 5\cos(\gamma)$$
and with the Theorem of sines we get
$$\frac{\sin(3\gamma)}{\sin(\gamma)}=\frac{c}{5}$$
With these equations you can eliminate $\gamma$ and then you can compute $c$
A: Let $AB=x$ and $AD$ be bisector of $\Delta ABC$.
Thus, $$\Delta ABD\sim\Delta CBA,$$ which gives
$$\frac{BD}{x}=\frac{x}{x+2}$$ or
$$BD=\frac{x^2}{x+2},$$ which gives
$$DC=x+2-\frac{x^2}{x+2}=\frac{4x+4}{x+2}$$ and since
$$\frac{AB}{AC}=\frac{BD}{DC},$$ we obtain:
$$\frac{x}{5}=\frac{\frac{x^2}{x+2}}{\frac{4x+4}{x+2}},$$
which gives $x=4$.
