Solution:

Let $AC=4.$ From this, we get $$CD=4,ED=3,BE=3.$$

Notice that $$\angle EDB = \angle EBD,$$$$\angle CAD = \angle CDA,$$ so $$\angle EDB + \angle CDA = 90 ^\circ.$$ Thus, $\angle EDC = 90^\circ.$ From the Pythagorean Theorem, we get $$EC = \sqrt{3^2+4^2}$$$$= \sqrt{25}$$$$=5.$$ Therefore, $BC = 8.$ This makes $$\tan BAC = \frac 84 = 2.$$

Now, we can get $$BD = 2\cdot 3\cos B$$$$= 6\sin A$$ and $$AD = 2\cdot 4\cos A$$$$= 8\cos A .$$ Thus, $$\frac{AD}{BD} = \frac{8 \cos A}{6\sin A}$$$$= \frac 4{3 \tan A}$$ $$=\frac 4{3\cdot 2}$$$$= \frac 23.$$

Thus, the answer is A .