Question

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:

$\tfrac12a+2b$

$\tfrac32b+\tfrac34a$

$2a-b$

$4b-\tfrac12a$

$a+b$

Answer: E

Concepts:angle bisector parallel lines isosceles triangle parallelogram

Difficulty rating: 2190

Small Hint:

Bisect angle $D$ and let the bisector meet $AB$ at $P$

Big Hint:

Angle chasing makes $\triangle APD$ isosceles, while $PBCD$ is a parallelogram

Solution:

Let the bisector of $\angle D$ meet $AB$ at $P.$ Because $\angle D=2\angle B,$ each half has measure $\angle B.$ Since $AB\parallel CD,$ the alternate interior angle $\angle APD$ also equals $\angle PDC.$ Thus triangle $APD$ is isosceles and $AP=AD=a.$ Also $PB\parallel CD$ and $BC\parallel PD,$ so $PBCD$ is a parallelogram. Hence $PB=CD=b.$ Therefore $AB=AP+PB=a+b.$

Therefore, the correct answer is E.

1970 AMC 12 Problem 30

Answer: E

Difficulty rating: 2190

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years