2. Trapezoid $ABCD$ with bases $AD$ and $BC$ is such that angle $ABD$ is a right angle and $BC + CD = AD$. Find the ratio of the bases $AD: BC$. (B. Obukhov)
Answer. $A D: B C=2$. First solution. On side $A D$, we mark off a segment $A E=B C$. Then $A B C E$ is a parallelogram, and $E D=C D$. Since $A B \| C E$, the diagonal $B D$, which is perpendicular to $A B$, is also perpendicular to $C E$. Therefore, it passes through the midpoint $F$ of the base $C E$ of the isosceles triangle $C D E$. Since $E F=C F, \angle E F D=\angle B F C$ and $\angle B C F=\angle F E D$, triangles $C F B$ and $E F D$ are congruent. Therefore, $E D=B C=A E$, from which the answer follows. Second solution. On side $A D$, we choose a point $K$ such that segment $B K$ is parallel to $C D$. Then $K D=B C$, and thus $A K=C D$. Moreover, $B K=C D$ (since $K B C D$ is a parallelogram), so $A K=B K$ and, therefore, the height $K H$ of triangle $A K B$ is also a median. Consequently, $K H$ is the midline of the right triangle $A B D$. Hence, $A D=2 K D=2 B C$.

Criteria. If the fact that a point on the hypotenuse, equidistant from the endpoints of a leg, is the midpoint of the hypotenuse is used without proof (but with a statement), 3 points are deducted. If an additional construction is made that can lead to a solution (for example, point $E$ or point $K$ mentioned in our solutions, or the extension of base $B C$ beyond point $C$ by a segment equal to $C D$, or the extension of side $D C$ beyond point $C$ by a segment equal to $B C$), but there is no further substantial progress, 1 point is awarded.