In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B$, and the measures of segments $AD$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to
$\text{(A) } \tfrac{1}{2}a+2b\quad \text{(B) } \tfrac{3}{2}b+\tfrac{3}{4}a\quad \text{(C) } 2a-b\quad \text{(D) } 4b-\tfrac{1}{2}a\quad \text{(E) } a+b$
With reference to the diagram above, let $E$ be the point on $AB$ such that $DE||BC$. Let $\angle ABC=\alpha$. We then have $\alpha =\angle AED = \angle EDC$ since $AB||CD$, so $\angle ADE=\angle ADC-\angle BDC=2\alpha-\alpha = \alpha$, which means $\triangle AED$ is isosceles. 
Therefore, $AB=AE+EB=a+b$, hence our answer is $\fbox{E}$.