The bisector of $\angle BAC$ of triangle $\Delta ABC$ cuts $BC$ at $D$ The bisector of $\angle BAC$ oF triangle $\Delta ABC$ cuts $BC$ at $D$ and circumcircle of triangle at $E$. if  
$$AD=5 \text{ cm} ,\ DE=3 \text{ cm},\ AC=4 \text{ cm}, $$
then what is the length of $AB$?
 A: 
Let $\angle ACB = \phi$. We have
\begin{align*}
\angle EBC &= \theta\\
\angle ABE &= 180^\circ -\theta - \phi
\end{align*}
Since the triangles $ABE$ and $ABC$ have the same circum circle, we have 
\begin{align}
\frac{8}{\sin(180^\circ - \theta - \phi)} = 2R = \frac{c}{\sin \phi} \label{eq1}
\end{align}
Also, in triangle $ACD$, $\angle ADC = 180^\circ - \theta - \phi$. Thus,
\begin{align}
\frac{5}{\sin \phi} = \frac{4}{\sin(180^\circ - \theta - \phi)} \label{eq2}
\end{align}
From the above two equations, we get 
\begin{align}
\frac{8}{c} = \frac{4}{5}
\end{align}
and hence $c=10$.
A: I've labeled $BD$ as $x$, $DC$ as $y$ and $AB$ as $c$.

There's probably a more efficient way to answer this, but....
By applying the chord theorem and the angular bisector theorem, you should be able to come up with two independent relationships among $x$, $y$ and $c$.
Next, apply the law of cosines to both $\Delta BAD$ and $\Delta DAC$, keeping in mind that $\angle BAD = \angle DAC = \theta$.
I was able to solve for $c$.  I got two potential solutions; one of them was not consistent with all the problem's givens, and can be ignored.
Good luck.
A: Join $\overline{BE}$. Using alternate segment theorem
$$\triangle(ACD) \equiv \triangle(AEB).$$
So, $$\frac{AC}{AD}= \frac{AE}{AB}.$$
Guess you are ok from here.
