Question about a trapezoid with three equal sides

Question

In a trapezoid the smaller base and two leges are equal to each other and the larger base is twice the smaller one. What is the largest angle of the trapezoid?

I know how to solve this problem. My question is about the following approach,

One can observe if we put three equal equilateral triangles adjacently together, we can form the desired trapezoid so the largest angle is $120^{\circ}$. But I'm wondering if it is possible to prove that the desired trapezoid is uniqe (i.e the trapezoid formed by these triangles is the only one which has the properties stated in the problem).

Answer 1

Here I could find a justification,

Assume the larger base is $AB$ and $G$ is its midpoint. The trapezoids wich have the legs equal to half of the larger base can be obtained by drawing a parallel line to $AB$ and its intersections with the circles are endpoints of the smaller base. Here we can see that the length of the smaller base can vary from $0$ to length of $AB$ (exclusively). So there is only one such trapezoid with smaller base equals to half of the larger one (which is the one that can be constructed by putting three equilateral triangles together).

Answer 2

Let $M$ be the midpoint of $AB$.

1. Since the trapezoid is isosceles $$\angle DAB \cong \angle ABC.$$
2. Hence $$\triangle AMD \cong \triangle MBC$$ by SAS criterion, and they are both isosceles triangles.
3. Since $DC\parallel AB$, $$\angle AMD \cong \angle MDC,$$ and $$\angle BMC \cong \angle MCB.$$
4. ASA criterion yields $$\triangle AMD \cong\triangle DMC.$$ In particular $$DM\cong CM \cong AD \cong AM.$$ And your conclusion easily follows.