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 $\angle AMD$? $\textbf{(A)}\ 15 \qquad$ $\textbf{(B)}\ 30 \qquad$ $\textbf{(C)}\ 45 \qquad$ $\textbf{(D)}\ 60 \qquad$ $\textbf{(E)}\ 75$
Solution
Draw $DE\bot CM$. The right triangles $DAM$ and $DEM$ share $DM$ and have $\angle AMD=\angle EMD$, so $\triangle DAM\cong \triangle DEM$ and $DE=DA=3$.
$AB\parallel CD$ gives $\angle DCE=\angle CMB$; with the right angles at $E$ and $B$ and $DE=CB=3$, $\triangle CDE\cong \triangle MCB$.
Hence $CM=CD=6$, and with $BC=3$, $\sin\angle BMC=\frac12$, so $\angle BMC=30^\circ$.
Rectangle $ABCD$ has $AB = \sqrt{2}$ and $BC = 1$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$?
