10.2. In a rectangular trapezoid $A B C D$, the sum of the lengths of the bases $A D$ and $\mathrm{BC}$ is equal to its height $\mathrm{AB}$. In what ratio does the bisector of angle $\mathrm{ABC}$ divide the lateral side $\mathrm{CD}$?
Answer: 1:1.

Solution. Let the intersection of the bisector of angle $\mathrm{ABC}$ with line $\mathrm{AD}$ be $\mathrm{Q}$. Since angle $\mathrm{ABC}$ is a right angle, triangle $\mathrm{ABQ}$ is a right isosceles triangle, and the length of segment $\mathrm{AQ}$ is equal to the length of the height $\mathrm{AB}$, which is the sum of the lengths of $\mathrm{AD}$ and $\mathrm{BC}$. Therefore, point $\mathrm{Q}$ is located on the extension of the base $\mathrm{AD}$ beyond point $\mathrm{D}$, and the length of segment $\mathrm{DQ}$ is equal to the length of the base $\mathrm{BC}$. Thus, quadrilateral $\mathrm{BCQD}$ is a parallelogram, and the bisector $\mathrm{BQ}$ of angle $\mathrm{ABC}$, which is a diagonal in it, bisects the side $\mathrm{CD}$, which is the other diagonal.

Grading criteria. It is proven that $\mathrm{Q}$ is located on the extension of the base $\mathrm{AD}$ beyond point $\mathrm{D}$ and the length of segment $\mathrm{DQ}$ is equal to the length of the base $\mathrm{BC}: 3$ points. If it is not proven that $\mathrm{Q}$ is located on the extension of the base $\mathrm{AD}$ beyond point $\mathrm{D}$: deduct 2 points.