imc

Let $\theta =\angle BAD=\angle DAC$. Notice $\tan \theta =BD$ and $\tan 2\theta =2$. By the double angle identity, $$2=\frac{2BD}{1-B{D}^2}⟹BD=\boxed{\frac{\sqrt{5}-1}{2}}.$$ Remarks: You could also use tangent half angle formula