5. Given a rectangle $A B C D$ whose side $A B$ is twice as long as side $B C$. A point $M$ is chosen on side $C D$ such that $\measuredangle A M D=\measuredangle A M B$.

a) Calculate $\measuredangle A M B$.

b) If the length of segment $D M$ is equal to 1, calculate the area of rectangle $A B C D$.
Solution. a) From $A B \| C D$ it follows that $\measuredangle \mathrm{BAM}=\measuredangle A M D$ (alternate interior angles), so $\measuredangle \mathrm{BAM}=\measuredangle \mathrm{AMB}$ (see diagram on the right). Therefore, triangle $A B M$ is isosceles and $B M=A M$. The hypotenuse $B M$ of the right triangle $B C M$ is twice the length of the leg (since by assumption $A B=2 B C$, and $B M=A B$).

As we know, such a triangle is half of an equilateral triangle, so $\measuredangle B M C=30^{\circ}$. Finally, $\measuredangle A M B+\measuredangle A M D=150^{\circ}$, hence $\measuredangle A M B=\measuredangle A M D=75^{\circ}$.

b) Let $B C=b$. Then $C D=2 b$. Since $C D=C M+D M$, and $D M=1$ and $C M=\sqrt{B M^{2}-B C^{2}}=b \sqrt{3}$, we get $2 b=b \sqrt{3}+1$, from which we find $b=\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}=2+\sqrt{3}$. Now the area of the triangle is

$$
P=2 b^{2}=2(2+\sqrt{3})^{2}=2(7+4 \sqrt{3})
$$