Problem 7. In an isosceles trapezoid $A B C D$ with bases $A D$ and $B C (A D > B C)$, the lateral side is 20 cm, and the angle $B A C$ is $45^{\circ}$. Let $O$ be the center of the circle circumscribed around $A B C D$. It turns out that the lines $O D$ and $A B$ are perpendicular. Find the length of the base $A D$ of the trapezoid.
Answer: $10(\sqrt{6}+\sqrt{2})$ cm.

Solution. Since $O$ - the center of the circumscribed circle - lies on the perpendicular bisector of segment $AB$, point $D$ also lies on this perpendicular bisector. Therefore, $AD = BD$, from which $\angle DAB = \angle DBA = \alpha$.

Due to the symmetry of the isosceles trapezoid, $\angle BDC = \angle BAC = 45^{\circ}$, and also $\angle CDA = \angle BAD = \alpha$. Thus, $\angle ADB = \angle CDA - \angle BDC = \alpha - 45^{\circ}$. From the sum of the angles in triangle $BAD$, we have: $\alpha + \alpha + (\alpha - 45^{\circ}) = 180^{\circ}$, i.e., $\alpha = 75^{\circ}$.

Finally, if we denote the midpoint of segment $AB$ as $M$, then from the right triangle $AMD$ we have $AD = \frac{AM}{\cos 75^{\circ}} = \frac{AB}{2 \cos 75^{\circ}}$. Further,

$$
\cos 75^{\circ} = \cos (45^{\circ} + 30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}
$$

i.e., $AD = \frac{AB(\sqrt{6} + \sqrt{2})}{2}$, from which the answer follows.