Finding t in basic geometric interpolation

Question

I have the following problem related to interpolation and was wondering if anyone had any ideas? In the following diagram, there exists known-points: A, B, C, D, E, unknown points F, G.

I'm trying to solve for: t, such that:

t = AF/AB = CG/CD

I'm a little lost here as I'm not sure of the best way to place points F and G to achieve this?

Thanks

Edit: To clarify, AB and CD are not necessarily parallel. In fact I tried this using the suggested approach when they are not parallel and obviously got a wrong result:

Is there a solution that can calculate the correct value for t when the edges aren't parallel?

Answer 1

Notation: $a$ is the vector from the origin to point $A$. Similarly for the other points.

The point $G$ can be written as $g=(1-t)c+td,\,0\leq t\leq 1$.

Similarly for point $F$ as $f=(1-s)a+sb,\,0\leq s\leq1$.

Since the three points are collinear, the line segments $\overline{EF}$ and $\overline{EG}$ must be parallel. This corresponds to the vectors $f-e$ and $g-e$ being parallel. You can show that this is done when the equation $$ \left|\begin{matrix}e_1&e_2&1\\f_1&f_2&1\\g_1&g_2&1\end{matrix}\right|=0 $$ is satisfied and $e=[e_1,\,e_2]^T, for example$. This equation can be solved for either $s$ or $t$ in terms of the other variable.

You will need one more piece of information to uniquely determine $F$ and $G$. At present, you have one free parameter, so that one of $F$ and $G$ can be at any point on a line, with the third satisfying the above equation and lying on the appropriate line.

Answer 2

Let $O$ be an intersection of lines generated by $AC$ and $BD$. Then line generated by $O$ and $E$ generates $F$ and $G$ (Tales theorem).