2010/11 AHSMC Part 1 Problem 9

Problem 9 of 16

In the quadrilateral $ABCD$ , $AB$ is parallel to $DC$ , $DC = 2AB$ , $\angle ADC = 30^\circ$ and $\angle BCD = 50^\circ$ . Let $M$ be the midpoint of $CD$ . The measure of $\angle AMB$ is

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Since $AB = DM$ and $AB$ is parallel to $DM$ , $ABMD$ is a parallelogram. Similarly, $ABCM$ is a parallelogram. Therefore, $\angle AMD = \angle BCD = 50^\circ$ and $\angle BMC = \angle ADC = 30^\circ$ . Therefore, $\angle AMB = 180^\circ - \angle AMD - \angle BMC = 180^\circ - 50^\circ - 30^\circ = 100^\circ$ . The answer is (C).

University of Alberta

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