jmc

Let $s$ be the length of a side of equilateral triangle $ABC$, and let $E$ be the foot of the perpendicular from $D$ to $\overline{BC}$. It follows that $△BDE$ is a $45-45-90$ triangle and $△CDE$ is a $30-60-90$ triangle. It follows that $BE=DE$ and $CE=DE/\sqrt{3}$, so $$s=BC=BE+EC=DE+DE/\sqrt{3}=DE\cdot \left(1+\frac{1}{\sqrt{3}}\right).$$It follows that $DE=\frac{s}{1+\frac{1}{\sqrt{3}}}=\frac{s}{\frac{\sqrt{3}+1}{\sqrt{3}}}=\frac{s\sqrt{3}}{1+\sqrt{3}},$ so $CE=DE/\sqrt{3}=\frac{s}{1+\sqrt{3}}$ and $CD=2CE=\frac{2s}{1+\sqrt{3}}$.

Since triangles $ADB$ and $CDB$ share the same height, it follows that the ratio of their areas is equal to the ratio of their bases, namely $AD/CD$. Since $AD=s-CD$, then $$\frac{AD}{CD}=\frac{s}{CD}-1=\frac{s}{\frac{2s}{1+\sqrt{3}}}-1=\frac{1+\sqrt{3}}{2}-1=\frac{\sqrt{3}-1}{2}.$$Thus, the ratio of the area of triangle $ADB$ to the area of triangle $CDB$ is $\boxed{\frac{\sqrt{3}-1}{2}}$.