Trapezoid $ABCD$ has $AD||BC$, $BD = 1$, $\angle DBA = 25^{\circ}$, and $\angle BDC = 50^{\circ}$. The ratio $BC: AD$ is $2: 1$. What is $CD$?

1. Let $\alpha=\angle ADB=\angle DBC$ and $x=\alpha+25^\circ$. By the Law of Sines, $AD=\dfrac{\sin25^\circ}{\sin x}$ and $BC=\dfrac{\sin50^\circ}{\sin(x+25^\circ)}$.
2. Using $BC/AD=2$ and $\sin50^\circ=2\sin25^\circ\cos25^\circ$ gives $\dfrac{2\cos25^\circ\sin x}{\sin(x+25^\circ)}=2$, hence $x=90^\circ$ and $\alpha=65^\circ$.
3. Since $BD=1$, $AD=\sin25^\circ$ and $BC=2\sin25^\circ$. Thus $CD=BC\dfrac{\sin65^\circ}{\sin50^\circ}=\dfrac{2\sin25^\circ\cos25^\circ}{2\sin25^\circ\cos25^\circ}=1$.