2011 AMC 12B Problems/Problem 10

Problem

Rectangle $ABCD$ has $AB=6$ and $BC=3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD$ . What is the degree measure of $\angle AMD$ ?

$\textrm{(A)}\ 15 \qquad \textrm{(B)}\ 30 \qquad \textrm{(C)}\ 45 \qquad \textrm{(D)}\ 60 \qquad \textrm{(E)}\ 75$

Solution

Since $AB \parallel CD$ , $\angle AMD = \angle CDM$ hence $CM=CD=6$ . Therefore $\angle BMC = 30^\circ$ . Therefore $\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}$

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2011 AMC 12B Problems/Problem 10

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