$\vec{AC}$ is the angle bisector of the angle between $\vec{AB}$ and $\vec{AD}$ which is $\frac{2\pi}{3}$,$15|\vec{AC}|=3|\vec{AB}|=5|\vec{AD}|$ In a quadrilateral $ABCD$,$\vec{AC}$ is the angle bisector of the angle between $\vec{AB}$ and $\vec{AD}$ which is $\frac{2\pi}{3}$,$15|\vec{AC}|=3|\vec{AB}|=5|\vec{AD}|$.If $\theta$ is the angle between $\vec{BA}$ and $\vec{CD}$,then prove that $\cos\theta=\frac{2}{\sqrt7}$

My Attempt:
In quadrilateral $ABCD$,$\angle BAC=\angle CAD=\theta=\frac{\pi}{3}$
In triangle $ABC,$apply cosine law,
$\cos\angle BAC=\frac{AC^2+AB^2-BC^2}{2AC\times AB}$

Putting $\frac{AC}{AB}=\frac{1}{5}$ and $\cos\angle BAC=\frac{1}{2}$
We get $\frac{BC}{AB}=\frac{\sqrt{21}}{5}$
Similarly applying cosine law in triangle $CAD$,we get
$\frac{DC}{AC}=\sqrt7$
But then i got stuck,please help me.Thanks.
 A: I have expresses every angle in degrees,since i am more comfortable with that.
Now I continue where you got stuck,so:
If you consider $\Delta ADC$ we have by Law of Sines that: $$\cfrac{DC}{ \sin 60^\circ }=\cfrac {AC}{\sin \angle ADC} $$
Since we know that $DC/AC=\sqrt 7$ ,we have that $DC=\sqrt{7} \cdot AC$
Hence,applying this to our previous equality ,we get:
$$\cfrac{ \sqrt {7} \cdot AC}{ \sin 60^\circ}=\cfrac {AC}{ \sin \angle ADC}$$
Simplifying this and rearranging for $\sin \angle ADC$,we have :
$$\sin\angle ADC =\cfrac { \sqrt {21}}{14}$$
$$\angle ADC =  \sin^{-1} \cfrac{ \sqrt {21}}{14}$$
Now ,if you consider $\Delta AXD $ where $X$ is the point of intersection of lines $AB$ and $DC$ (hence the angle intercepted by these lines  is your $\theta$ ),we have that $$\angle DAX + \angle ADC + \theta = 180^\circ$$
$$\theta =60^\circ - \angle ADC=60^\circ - \sin^{-1} \cfrac{ \sqrt {21}}{14}$$
$$\cos \theta = \cos (60^\circ - {\sin^{-1} \cfrac {\sqrt {21}}{14}})=0,755928946...=\cfrac{2}{\sqrt{7}}$$

