Consider the figure. It is supposed to be a tennis court. A ball is served at $F$. It's trajectory is a straight line. The ball touches the ground at $A$. Find the distance $\mathbf x'$ from $A$ to the net given that $FE=7ft$ and the distance $ED=39ft$. Also we know that $BC=3ft$.
I have no arguments to say that $\dfrac{7'}{39}=\dfrac{3'}{x'}$. I can see that $\triangle EDC\sim\triangle AXC$ where $X$ is the point of intersection of a perpendicular from $A$ to $\overleftrightarrow{DC}$. And that $\triangle FEA \sim \triangle BCA$.
I have the following idea: all the possible trajectories of a ball thrown from $F$ such that the ball touches the net lie on a plane, namely the plane passing through the net and the point $F$. This plane intersects the tennis court in a line $r$ which is parallel to the net and is at a distance $d=AX$ from it: this is the same as saying that every ball thrown from $F$ will land at the same distance from the net. So you can consider the trajectory of a ball thrown parallel to the side of the court, touching the net on a point $D'$ directly over the point $D$. The ball wil land on a point $A'$, which will be distant $d$ from the net. This reduces the problem to a $2D$ problem, and using similarity you can see that \begin{equation} d:3=39:(7-3) \end{equation} so \begin{equation} d=117/4 \end{equation}