In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $$\angle BAC = 45^{\circ}$$ and $$\angle ABC=\theta\ $$, then $$\frac{AD}{BE}$$ equals

A

$$\sqrt{2} \cos \theta$$

B

$$\frac{(\sin \theta + \cos \theta)}{\sqrt{2}}$$

C

1

D

$$\sqrt{2} \sin \theta$$