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Find AB in the diagram

Source: 1970 AHSME Problem 30

April 16, 2014
trigonometrygeometryparallelogramtrapezoidAMC

Problem Statement

In the accompanying figure, segments ABAB and CDCD are parallel, the measure of angle DD is twice the measure of angle BB, and the measures of segments ABAB and CDCD are aa and bb respectively. Then the measure of ABAB is equal to
<spanclass=latexbold>(A)</span>12a+2b<spanclass=latexbold>(B)</span>32b+34a<spanclass=latexbold>(C)</span>2ab<spanclass=latexbold>(D)</span>4b12a<spanclass=latexbold>(E)</span>a+b<span class='latex-bold'>(A) </span>\dfrac{1}{2}a+2b\qquad<span class='latex-bold'>(B) </span>\dfrac{3}{2}b+\dfrac{3}{4}a\qquad<span class='latex-bold'>(C) </span>2a-b\qquad<span class='latex-bold'>(D) </span>4b-\dfrac{1}{2}a\qquad <span class='latex-bold'>(E) </span>a+b
[asy] size(175); defaultpen(linewidth(0.8)); real r=50, a=4,b=2.5,c=6.25; pair A=origin,B=c*dir(r),D=(a,0),C=shift(b*dir(r))*D; draw(A--B--C--D--cycle); label("AA",A,SW); label("BB",B,N); label("CC",C,E); label("DD",D,S); label("aa",D/2,N); label("bb",(C+D)/2,NW); //Credit to djmathman for the diagram[/asy]