Problem 3. Given a rectangle $A B C D$ where $\overline{A B}=2 \overline{B C}$. On the side $C D$, a point $M$ is chosen such that the angle bisector of $\measuredangle D M B$ passes through vertex $A$. Determine the measure of $\measuredangle A M B$.
Solution. Let $\overline{A B}=a, \overline{B C}=b$. Then $\overline{A B}=2 b$. Since $A M$ is the angle bisector of $\measuredangle D M B$, we get $\measuredangle D M A=\measuredangle A M B$. On the other hand, $\measuredangle D M A=\measuredangle B A M$, as alternate interior angles. Therefore, $\measuredangle B A M=$

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$\measuredangle A M B$, which means that triangle $A M B$ is isosceles and $\overline{B M}=\overline{A B}=2 b$. Therefore, in the right triangle $B M C$, one leg is $\overline{B C}=b$, and the other is $\overline{B M}=2 b$, so it is half of an equilateral triangle with side $\overline{B M}=2 b$. From this, we get $\measuredangle C B M=60^{\circ}$, which means that $\measuredangle B M C=30^{\circ}$. Therefore, $\measuredangle D M B=180^{\circ}-\measuredangle B M C=180^{\circ}-30^{\circ}$, so $\measuredangle A M B=\frac{1}{2} \measuredangle D M B=\frac{150^{\circ}}{2}=75^{\circ}$.