$\vec{A},\vec{B},\vec{C}$ and $\vec{D}$ are unit vectors ($|A|=1,|B|=1,|C|=1$ and $|D|=1$). The angle between the vectors,
1) $\vec{A}$ and $\vec{B}$ is $\theta_{1}$ ($\vec{A}.\vec{B}=\cos\theta_{1}$)
2) $\vec{B}$ and $\vec{C}$ is $\theta_{2}$ ($\vec{B}.\vec{C}=\cos\theta_{2}$)
3) $\vec{C}$ and $\vec{D}$ is $\theta_{3}$ ($\vec{C}.\vec{D}=\cos\theta_{3}$)
4) $\vec{D}$ and $\vec{A}$ is $\theta_{4}$ ($\vec{D}.\vec{A}=\cos\theta_{4}$) and
5) $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}=2\pi$.
The objective is to make a relationship between $\theta$'s (As an example $\theta_{1}+\theta_{2}=\pi$). Also there is one relationship with the vectors.
\begin{equation} (p+q+r)(\vec{A}+\vec{B}) + (p+q)\vec{C} +r\vec{D}=0 \end{equation}
$p,q,r$ are constants. How would I do that, So far I worked like this,
I take the dot product of vector $(\vec{A}-\vec{B})$ (because $(\vec{A}-\vec{B}).(\vec{A}+\vec{B})=0$) with above equation,
\begin{eqnarray} (p+q+r)(\vec{A}+\vec{B}).(\vec{A}-\vec{B}) + (p+q)\vec{C}.(\vec{A}-\vec{B}) +r\vec{D}.(\vec{A}-\vec{B})=0 \\ (p+q)\vec{C}.(\vec{A}-\vec{B}) +r\vec{D}.(\vec{A}-\vec{B})=0 \end{eqnarray}
Is that possible to reduce above equation in terms of $\theta$'s?