I don't have the best figure but this is all I got.
Length of $CD=l_1$ and of $AB=l_2$
$CD||AB $
Perpendicular distance between them$=d$
Therefore $$CE=DF=d$$
Angle subtended by normal and $A$ at any point on $CD$ is $\phi$ and by normal and $B$ is $\theta$
Therefore,
$$\angle ACE=\phi_1 \ \angle BCE=\theta_1$$$$
\angle ADF=\phi_2 \ \angle BDF=\theta_2$$
Let distance moved on $CD$ in directions $C\rightarrow D=x $ and distance of foot of normal on $ AB $ in direction $A\rightarrow B=y$
Now this is part of a problem I encountered in integration: $$\int_0^{l_1}k\left(\sin \phi+\sin\theta\right)dx$$
This meant I had to find a relation between $x$ and $y$ or $x$ and $(\phi,\theta).$ My geometry failed me.
Please Help.