In a parallelogram $ABCD, |\overrightarrow{AB}|=a;|\overrightarrow{AD}|=b$ and $|\overrightarrow{AC}|=c$, then $\overrightarrow{DB}.\overrightarrow{AB}$ has the value :

Question

$\begin{array}{1 1} \frac{1}{2}(a^2-b^2+c^2) \\ \frac{1}{4}(a^2+b^2-c^2) \\ \frac{1}{3}(b^2+c^2-a^2)\\ \frac{1}{2}(a^2+b^2+c^2) \end{array}$