I found a proof for this problem as follows:
Let $r$ be the distance marked with one stroke in the problem diagram and for angle $\alpha$ make the following construction:
The circles centered at A and B have radius $r$. Call the length of segment CF $f(\alpha)$. We desire that it be equal to $r$.
In the interval $\alpha_{min}<\alpha<\frac 23\pi $, where $\alpha_{min}$ is the angle when EC and CB are parallel, $f$ is monotonically decreasing with one limit at $0$ and the other at infinity. Therefore $f$ takes every positive real value, particularly $r$ exactly once. Now going back to the problem diagram: there's only one triangle ABC with the desired side lengths (unless $r$ is greater than the side of the inner triangle, when there are none), and it can easily be verified that there exists one solution with ABC being equilateral. Therefore that only solution.
Can this approach be used for other problems? And does it have a name?