Label the trapezoid as the shown below. The base of the quadrilateral is $\overline{AD}$. The lengths of the sides $\overline{AB},\overline{BC},\overline{CD},\overline{AD}$ are respectively $b,c,d$, and $a$. The diagonals $\overline{AC}$ and $\overline{BD}$ both have length $r$. Angles $\angle BAC$ and $\angle CAD$ are both $\alpha$, while $\angle ADB$ and $\angle BDC$ are both $\beta$.
Now because of how we define the "inside" of the quadrilateral, it must be the case that $2\alpha<\pi$ and $2\beta<\pi$. Furthermore, since $\overline{AB}$ and $\overline{BC}$ must intersect, $2\alpha+\beta<\pi$. Likewise, $\alpha+2\beta<\pi$.
This valid region for $\alpha$ and $\beta$ is shown below, excluding the boundary.
Note that because of this range of allowable values of $\alpha$ and $\beta$, the sines and cosines of these angles are positive.
Now, by equating expressions for the positions of points $B$ and $C$, we get the following equations.
$$\begin{align*}
r\cos\alpha&=a-d\cos2\beta & &(1)\\
b\cos2\alpha&=a-r\cos\beta & &(2)\\
r\sin\alpha&=d\sin2\beta & &(3)\\
a\sin2\alpha&=r\sin\beta & &(4)\\
\end{align*}$$
Rearranging $(1)$
$$a-r\cos\alpha=d\cos2\beta$$
Dividing by $(3)$
$$\frac{a-r\cos\alpha}{r\sin\alpha}=\frac{d\cos2\beta}{d\sin2\beta}$$
Solving for $r$
$$a-r\cos\alpha=r\sin\alpha\cot2\beta$$
$$r=\frac{a}{\cos\alpha+\sin\alpha\cot2\beta}$$
Solving $(2)$ and $(4)$ for $r$
$$r=\frac{a}{\cos\beta+\sin\beta\cot2\alpha}$$
Combining these equations
$$\cos\alpha+\sin\alpha\cot2\beta=\cos\beta+\sin\beta\cot2\alpha$$
Multiplying by $\sin2\alpha\sin2\beta$
$$\sin2\alpha\sin2\beta\cos\alpha+\sin2\alpha\cos2\beta\sin\alpha\\=\sin2\alpha\sin2\beta\cos\beta+\cos2\alpha\sin2\beta\sin\beta$$
Applying double angle theorems
$$4\sin\alpha\cos^2\alpha\sin\beta\cos\beta+(1-2\sin^2\beta)2\sin^2\alpha\cos\alpha\\=4\sin\alpha\cos\alpha\sin\beta\cos^2\beta+(1-2\sin^2\alpha)2\sin^2\beta\cos\beta$$
Distributing and factoring
$$4\sin\alpha\cos\alpha\sin\beta\cos\beta(\cos\alpha-\cos\beta)-4\sin^2\alpha\sin^2\beta(\cos\alpha-\cos\beta)\\+2(1-\cos^2\alpha)\cos\alpha-2(1-\cos^2\beta)\cos\beta=0$$
$$4\sin\alpha\cos\alpha\sin\beta\cos\beta(\cos\alpha-\cos\beta)-4\sin^2\alpha\sin^2\beta(\cos\alpha-\cos\beta)\\+2(\cos\alpha-\cos\beta)-2(\cos^3\alpha-\cos^3\beta)=0$$
$$(4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+2\\-2(\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta))(\cos\alpha-\cos\beta)=0$$
Now
$$(4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+2-2(\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta))<0$$
for all $\alpha$ and $\beta$ in the legal range†. (It is exactly zero at $\alpha=0,\beta=\frac{\pi}{2}$ and vice versa.) This means that
$$\cos\alpha-\cos\beta=0$$
$$\Rightarrow \alpha=\beta$$
Now clearly the base angles $2\alpha$ and $2\beta$ are equal. By SAS congruence, we get that $\triangle ABD$ and $\triangle DCA$ are congruent. Therefore the base side lengths $b$ and $d$ are equal.
†Verified graphically. A full proof would likely include function maximization to show that the expression does not exceed $0$.
