11.2. In the described quadrilateral $A B C D$, the measures of angles CDA, DAB, and ABC are $90^{\circ}, 120^{\circ}$, and $120^{\circ}$ respectively, and the length of side BC is 1 cm. Find the length of side AB.
Answer. $A B=2-\sqrt{3}$

Solution. Let the length of side AB be $x$. Extend sides DA and CB until they intersect at point E. Triangle EAB is isosceles with angles of 60 degrees at the base AB, making it equilateral. Triangle ECD is a right triangle with a 30-degree angle at vertex C, so $E D=\frac{1}{2} E C=\frac{x+1}{2}, D C=\frac{\sqrt{3}}{2} E C=\frac{\sqrt{3}(x+1)}{2}$. Therefore, $A D=E D-A E=E D-x=\frac{1-x}{2}$. Since quadrilateral ABCD is cyclic, $A D+B C=A B+C D \Leftrightarrow \frac{3-x}{2}=x+\frac{\sqrt{3}(x+1)}{2} \Leftrightarrow 3-3 x=\sqrt{3}(x+1)$, from which $x=2-\sqrt{3}$.

Grading criteria. ($\cdot$) Lengths of side CD and segment DE are expressed in terms of the length of AB: 2 points. ($\cdot$) Length of side AD is expressed in terms of the length of AB: 1 point. ($\cdot$) The condition for the cyclic quadrilateral ABCD is written: 1 point.