Let the eye be at $\displaystyle (0,0,h)$, and the lines be on the $\displaystyle xy$ plane and parallel to the $\displaystyle x$ axis, each at distance s from the $\displaystyle x$-axis. Let the eye be facing towards the positive $\displaystyle x$-axis. Let the plane $\displaystyle D$ be at distance $\displaystyle d$ from the eye. Let each point on that plane have coordinates $\displaystyle (u,v)$ (they are analogous to $\displaystyle (y,z)$ actually).

Here is a more informal derivation. The lines intersect the plane $\displaystyle D$ at $\displaystyle (d,s,0)$ and $\displaystyle (d,-s,0)$. Since the vanishing point is directly straight ahead, the projections of these lines on the plane $\displaystyle D$ must intersect at the point $\displaystyle (d,0,h)$. Since the lines are on a plane and we are projecting them on another plane, their projections must be lines also. Thus we have that the projections of these lines on plane $\displaystyle D$ are the lines starting at $\displaystyle (s,0)$, $\displaystyle (-s,0)$ and intersecting at $\displaystyle (0,h)$ in $\displaystyle (u,v)$ coordinates. From these you can use simple trig to deduce that the angle between these two projections must be $\displaystyle 2\cos^{-1}\left(\frac{s}{\sqrt{h^s+s^2}}\right)$ on the plane.

Surprisingly, this angle does not depend on the distance we have chosen for the plane. It seems counter-intuitive.

Here is the more complicated derivation that uses some vectors. The idea is to write down an equation of the line going from the eye at $\displaystyle (0,0,h)$ to a point on one of the lines $\displaystyle (p,s,0)$. Then we need to find the intersection of this line with the plane $\displaystyle D$. This will be at coordinates $\displaystyle (d,u,v)$.

To get the equation of a line, we need a position vector $\displaystyle \vec{R_o}$ and a direction vector $\displaystyle \vec{V}$, which we'll multiply by a parameter $\displaystyle t$ to get the vector equation for the line to be $\displaystyle \vec{R_o}+t\vec{V}$. So, choose the position vector $\displaystyle \vec{R_o} = <p,s,0>$ and the direction vector be from that point to the eye. By the rules of vector addition, this is $\displaystyle \vec{V}=<0,0,h>-\vec{R_o} = <-p,-s,h>$. Thus the vector equation for the line is

$\displaystyle \displaystyle{\vec{R_o}+t\vec{V} = <p(1-t),s(1-t),th>}$

Since we want the projection onto the plane $\displaystyle D$, we need to have the x-coordinates match, which means $\displaystyle p(1-t) = d$. From this we get that $\displaystyle (1-t) = d/p$ and $\displaystyle t = 1-d/p$. Subbing both of those in the equation of the line, we see that the intersection point is $\displaystyle (d,\frac{sd}{p},(1-\frac{d}{p})h)$

Now, if we set $\displaystyle p = d$, that is we are looking at the intersection between the line and the plane $\displaystyle D$, we will simply get $\displaystyle (d,s,0)$. However, the interesting part is when we let it go to infinity.

$\displaystyle \displaystyle{\lim_{p\to \infty}(d,\frac{sd}{p},(1-\frac{d}{p})h) = (d,0,h)}$

This is the same result as before! Only this time we have it more rigorously. Furthermore, since the coordinates of the projection of the line on the plane $\displaystyle D$ are $\displaystyle (d,\frac{sd}{p},(1-\frac{d}{p})h)$, then in $\displaystyle (u,v)$ coordinates, this is just $\displaystyle (\frac{sd}{p},(1-\frac{d}{p})h)$. These are just parametric equations of the line in parameter $\displaystyle p$. They form a system of parametric equations

$\displaystyle \displaystyle{u = \frac{sd}{p}}$

$\displaystyle \displaystyle{v = (1-\frac{d}{p})h) }$

From these we get that $\displaystyle d/p = u/s$ and $\displaystyle v = h - \frac{h}{s}u$, which is the equation of the projection of the line onto the plane $\displaystyle D$. You can see that it is a line! the other line has equation $\displaystyle v = h + \frac{h}{s}u$. With trig you can get that the angle between one line and the vertical axis is $\displaystyle \theta = \cos^{-1}\left(\frac{s}{\sqrt{h^s+s^2}}\right)$ and the angle between the lines is just $\displaystyle 2\theta$

Here is the reason why I think this doesn't depend on the distance $\displaystyle d$ of the plane $\displaystyle D$. Say we are looking at two parts of the line. One at distance $\displaystyle p_1$ and the other at distance $\displaystyle p_2$. The projections on the plane $\displaystyle D$ will BOTH involve the parameter $\displaystyle d$, but since we are looking at the relation between those two distances $\displaystyle p_1$ and $\displaystyle p_2$, we will have a sort of cancellation of the parameter $\displaystyle d$. I think this is a neat result.