+421 We have a 30 60 90 triangle.
Let $BC = x.$ $AC = x \sqrt{3},$ and $AB = 2x.$
We also have $AC * CB = \sqrt{3} * AB.$
Let $DB = y.$ Then $$y^2+3=x^2,$$ and $$3+(2x-y)^2 = (x \sqrt {3})^2.$$ (by the pythagorean theorem.)
Expanding, we have $$y^2+3=x^2,$$ and $$4 x^2 - 4 x y + y^2 + 3 = 3 x^2.$$
If we solve this equation, we get $x=2$ and $y=1,$ for integer solutions. Thus $BD = y = 1.$
