Homothety of general figures

Question

Is there an equivalent statement (other than finding the center of homothecy) which assures homothecy? I have the following problem. I have two figures in a plane, $A$ and $B$ (both convex), translated so that they are touching the $y$-axis and are in the first quadrant. The projection of the figure $A$ on the $x$-axis is the interval $[0,a]$, and projection of the figure $B$ is the interval $[0,b]$. Denote $\alpha(t)=at$ and $\beta(t)=bt$ for some $t\in[0,1]$. Also denote Also denote $f(x)$ to be the length of the cross section of figure $A$ at coordinate $x$, same for $B$ and $g(x)$. In the problem which I was doing I was able to prove that $\dfrac{f(\alpha(t))}{g(\beta(t))}=\sqrt{\dfrac{S(A)}{S(B)}}$, for all $t$, where $S$ denotes area. This itself isn't enough for homothecy (counterexample would be a parallelogram and a rectangle with an equal side). However, I can reach the same conclusion if I rotate both figures by the same angle. Is this enough to claim they are homothetic?