1970 AHSME Problems/Problem 30

Problem

$[asy] draw((0,0)--(2,2)--(5/2,1/2)--(2,0)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(5/2,1/2),SE);MP("D",(2,0),S); MP("a",(1,0),N);MP("b",(17/8,1/8),N); [/asy]$

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

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

Solution

$\fbox{E}$

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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1970 AHSME Problems/Problem 30

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