Answer: E

Small Hint:

Because $AC=CD,$ triangle $ACD$ is isosceles

Big Hint:

Express $\angle CDA$ using $\angle DAB$ and $\angle ABC$

Solution:

Let $\alpha=\angle CAB,$ $\beta=\angle ABC,$ and $x=\angle BAD.$ Since $D$ lies on $BC,$ $$\angle ADC=x+\beta$$ as an exterior angle of triangle $ABD.$ Because $AC=CD,$ triangle $ACD$ is isosceles, so $\angle CAD=\angle ADC.$ But $\angle CAD=\alpha-x.$ Hence $$\alpha-x=x+\beta,$$ so $2x=\alpha-\beta=30^\circ$ and $x=15^\circ.$

Thus, the correct answer is E.