Set $\angle DBA=\alpha$ and compare triangles $ABC$ and $ABD$

Use the law of sines to obtain an equation involving $\sin5\alpha,$ $\sin2\alpha,$ and $\sin\alpha$

Let $\angle DBA=\alpha.$ Then $\angle DBC=\angle CAB=2\alpha,$ so in $\triangle ABC$ the angles are $2\alpha,$ $3\alpha,$ and $180^\circ-5\alpha.$ Because $AD\parallel BC,$ triangle $ABD$ has angles $\alpha,$ $2\alpha,$ and $180^\circ-3\alpha.$

Applying the law of sines in the two triangles and using $AD=BC$ gives $$\frac{AB}{BC}=\frac{\sin5\alpha}{\sin2\alpha}=\frac{\sin2\alpha}{\sin\alpha}.$$ With $u=\cos^2\alpha,$ this becomes $$16u^2-16u+1=0.$$ Since $5\alpha<180^\circ,$ the valid root is $u=\frac{2+\sqrt3}{4}=\cos^2 15^\circ,$ so $\alpha=15^\circ.$ Thus $\angle ACB=105^\circ.$ In $\triangle AOB,$ the angles at $A$ and $B$ are $30^\circ$ and $15^\circ,$ so $\angle AOB=135^\circ.$ Therefore $$\begin{aligned}r&=\frac{105}{135}=\frac79,\\\left\lfloor1000r\right\rfloor&=777.\end{aligned}$$