One case is $\boxed{s=r}$

Attachments:

$\angle A = \frac{\pi}{3}$ or $\frac{\pi}{5}$ Let $r=1$ Obvious , that $ABCD$ is trapezoid. $BD^2=AB^2+AD^2-2AB*AD cos A=s^2+(s+1)^2-2s(s+1)cos A$ $BD^2=BC^2+CD^2-2BC*CD cos C=s^2+s^2-2s^2cos C$ $cos C=-cos A$ $s^2+(s+1)^2-2s(s+1)cos A=2s^2+2s^2cosA$ $cosA(2s^2+2s(s+1))=2s+1$ $cos A= \frac{2s+1}{2s(2s+1)}=\frac{1}{2s}$ $BD^2=2s^2+2s^2* \frac{1}{2s}=s(2s+1)$ $BD^2=4r^2 sin^2 A=4(1-cos^2 A)$ $s(2s+1)=4-\frac{1}{s^2}$ $s(2s+1)=\frac{(2s-1)(2s+1)}{s^2}$ $s=\frac{2s-1}{s^2}$ $s^3-2s+1=0 $ $s=1, \frac{\sqrt{5}-1}{2}$ $cos A=\frac{1}{2}, \frac{\sqrt{5}+1}{4}$ $A= \frac{\pi}{3}, \frac{\pi}{5}$

Sketch: Use Law of Sines twice on the appropriate triangles, that then gives \[ \frac12 = \sin(3 \theta) - \sin \theta \]where $\theta := \angle BAC$. Note that $\theta$ is sufficient to find all the needed angles. Now the above equation is easily solved by using the addition theorems of Sines and Cosines, as that gives us a polynomial equation in degree $3$ and it's easy to guess one answer. After that you still have to show that it's there are indeed quadrilaterals with such angles.