Prove that the image of a line lies on a plane

Question

Problem I have:
There's a point E that is the center of projection. There is a plane s such that it does not contain E. Let L be a line such that it does no go through E, and does not lie in S. Prove that the image of L is in S.

I have tried to solve this using properties of similar triangles and also some vector stuff but I haven't gotten any closer to the proof.
Any Ideas on how I can prove this?

--Edit--
Regarding what I've tried, we know that a line lies on a plane if two points on the line also lie on the plane.
To do this, I assumed E is at (0,0,0) and that L, and S intersect the X-Axis at (x1, 0, 0) and (x, 0, 0) respectively.
I also assumed that a line that goes through the origin intersects the S and L at (x, y, z) and (x1, y1, z1) respectively.
This gives me two similar triangles (that I thought will help with the proof) and points (to make equations for the lines) but none of that led to a proof and I have no idea how to proceed.