Finding an angle in a triangle, given the angle bisector and some conditions. 

$ABC$ is a triangle in which $\angle B= 2\angle C$. $D$ is a point on side $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$. 
What is the measure of $\angle BAC? $ 
I tried using angle bisector theorem, similarity, but the sides, don't include the angles. Also I got a relation, $3x+2\theta=180^{\circ}$, where $\angle ACB=x;\angle BAD=\theta$. 
But I cannot derive one more relation, please help. 
Please don't use trigonometry, nor any constructions. We have to do without them, simply.

Notice, $$\angle ADB=180^\circ -\left(\angle B+\frac{\angle BAC}{2}\right)$$
$$\angle B+\angle C+\angle BAC=180^\circ\iff 2\angle C+\angle C+\angle BAC=180^\circ$$ $$\iff \angle C=\frac{180^\circ-\angle BAC}{3}\tag 1$$$$\iff \angle B=\frac{2(180^\circ-\angle BAC)}{3}\tag 2$$
apply sine rule in $\triangle ABD$ $$\frac{\sin\angle B}{AD}=\frac{\sin \angle ADB}{AB}$$
$$\frac{\sin\angle B}{AD}=\frac{\sin \left(\angle B+\frac{\angle BAC}{2}\right) }{AB}\tag 3$$ Similarly, apply sine rule in $\triangle ADC$
$$\frac{\sin\angle C}{AD}=\frac{\sin \left(\frac{\angle BAC}{2}\right)}{CD}$$
Setting $CD=AB$, we get 
$$\frac{\sin\angle C}{AD}=\frac{\sin \left(\frac{\angle BAC}{2}\right)}{AB}\tag 4$$ 
Dividing (3) by (4), we get 
$$\frac{\sin\angle B}{\sin \angle C}=\frac{\sin \left(\angle B+\frac{\angle BAC}{2}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$
Setting the values of $\angle B$ & $\angle C$, we get 
$$\frac{\sin \left(\frac{2(180^\circ-\angle BAC)}{3}\right)}{\sin \left(\frac{180^\circ-\angle BAC}{3}\right)}=\frac{\sin \left(\frac{2(180^\circ-\angle BAC)}{3}+\frac{\angle BAC}{2}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$
$$2\cos \left(60^\circ-\frac{\angle BAC}{3}\right)=\frac{\sin \left(120^\circ-\frac{\angle BAC}{6}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$
I hope you can solve for $\angle BAC$
