Problem 2. Given is the rectangle $ABCD$ where the length of side $AB$ is twice the length of side $BC$. On side $CD$, a point $M$ is chosen such that the angle $AMD$ is equal to the angle $AMB$.

a) Determine the angle $AMD$.

b) If $\overline{DM}=1$, what is the area of rectangle $ABCD$?
Solution. Since $A B$ and $C D$ are parallel, $\measuredangle B A M=\measuredangle A M D$ (alternate interior angles). From the condition $\measuredangle A M B=\measuredangle A M D$, we conclude that $\measuredangle B A M=\measuredangle A M B$. Therefore, triangle $A M B$ is isosceles and $\overline{A B}=\overline{M B}$. In triangle $B C M$

![](https://cdn.mathpix.com/cropped/2024_06_05_028d7bfd550280da1c84g-10.jpg?height=236&width=387&top_left_y=1223&top_left_x=1056)
the hypotenuse $B M$ is twice the length of the leg $B C (\overline{A B}=2 \overline{B C}$, and $\overline{A B}=\overline{M B})$. Then $\measuredangle B M C=30^{\circ}$, so

$$
\measuredangle A M B+\measuredangle A M D=150^{\circ}
$$

From this, we get

$$
\measuredangle A M B=\measuredangle A M D=75^{\circ}
$$

Let $\overline{B C}=b, \overline{C D}=2 b$. From the condition $\overline{D M}=1$ we have

$$
\overline{C D}=\overline{C M}+\overline{M D}=\overline{C M}+1 \text {, i.e., } 2 b=\overline{C M}+1
$$

From the Pythagorean theorem, it follows that

$$
\overline{C M}=\sqrt{\overline{B M}^{2}-\overline{B C}^{2}}=\sqrt{(2 b)^{2}-b^{2}}=b \sqrt{3}
$$

Accordingly, $2 b=b \sqrt{3}+1$, i.e., $b=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$. The area of the rectangle is $P=a b=2 b^{2}=2(2+\sqrt{3})^{2}=2(7+4 \sqrt{3})$.