AHSME 1970 Problem 30

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 $\text{(A) } \tfrac{1}{2}a+2b \text{(B) } \tfrac{3}{2}b+\tfrac{3}{4}a \text{(C) } 2a-b \text{(D) } 4b-\tfrac{1}{2}a \text{(E) } a+b$