Given a parallelogram $AB\Gamma\Delta$ and a line $(\epsilon)$ as shown in the figure below prove that $(AKB)(B\Lambda\Delta)(\Gamma M\Delta)(\Gamma N A)=1$
Notation: $(ABC)$ denotes the (signed) ratio that $B$ splits the segment $(AC)$ for three collinear points $A,B,C$ (i.e $(ABC)=\lambda \Leftrightarrow \vec{OB} = \frac{1}{\lambda+1}\vec{OA} + \frac{\lambda}{\lambda+1}\vec{OC}$)
Attempt: I tried applying Menelaus's theorem on the triangles $\triangle AB\Delta$ and $\triangle A\Gamma\Delta$. This would give $(AKB)(B\Lambda\Delta)(\Delta P A)=-1 $ as well as $(AN\Gamma)(\Gamma M \Delta)(\Delta P A)=-1$. But from there I can't seem to be able to continue. If instead of $(\Gamma M \Delta)$ we would have $(\Delta M \Gamma)$ then by dividing these two relations we would get the required result since $(ABC)=\frac{1}{(CBA)}$