Answer
BD=5
AB=10
Work Step by Step
We let BD=x and AB=2x. Applying the Pythagorean theorem:
$$(6^2)+(x+3)^2=(2x)^2 \\ x^2+6x+45=4x^2 \\ x_{1,2}=\frac{-6\pm \sqrt{6^2-4(-3)\cdot \:45}}{2(-3)} \\ x=-3, 5$$
x can't be negative, so $x=5$
Thus:
$$ BD=5 \\ AB=(2)(5)=10$$
