5. (20 points) In an acute-angled triangle $\mathrm{ABC}$, a point $\mathrm{D}$ is chosen on side $\mathrm{BC}$ such that $\mathrm{CD}: \mathrm{DB}=2: 1$, and a point $\mathrm{K}$ is chosen on segment $\mathrm{AD}$ such that $\mathrm{AK}=\mathrm{CD}+\mathrm{DK}$. A line is drawn through point $\mathrm{K}$ and vertex $\mathrm{B}$, intersecting side $\mathrm{AC}$ at point $\mathrm{E}$. Triangle $\mathrm{AEK}$ is isosceles ($\mathrm{AE}=\mathrm{EK}$). Find the measure of angle $\mathrm{ADC}$ in degrees.
Solution: Extend SV beyond point V so that BN = BD. Draw NM || BE. NM intersects AD at point L. Draw segment $\mathrm{CH} \perp \mathrm{AD}$. Extend it to $\mathrm{P}$ such that $\mathrm{HP}=\mathrm{HC}, \mathrm{PN}|| \mathrm{AD}$.

In triangle DLN, segment $\mathrm{BK}$ is the midline, therefore, $\mathrm{DK}=\mathrm{KL}$ and thus $\mathrm{AL}=\mathrm{AK}-\mathrm{LK}=\mathrm{AK}-\mathrm{DK}=\mathrm{CD}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9144fd154fefa9836059g-18.jpg?height=631&width=805&top_left_y=1809&top_left_x=777)

Triangle САР is isosceles (AH is the median and altitude), therefore $\angle \mathrm{PAD}=\angle \mathrm{DAC}$, but $\angle \mathrm{DAC}=\angle \mathrm{AKE}$ (by condition), $\angle \mathrm{AKE}=\angle \mathrm{ALM}$ (corresponding), $\angle \mathrm{ALM}=\angle \mathrm{NLD}$ (vertical), $\angle \mathrm{NLD}=\angle \mathrm{NLP}$ (alternate interior), hence, LN || AP, so ALNP is a parallelogram.

In the right triangle $\mathrm{CPN}$, the length $\mathrm{CN}=2 \mathrm{CD}=2 \mathrm{AL}=2 \mathrm{PN}$, i.e., $\angle \mathrm{PCN}=30^{\circ}$. Therefore, $\angle \mathrm{ADC}=60^{\circ}$.

Answer: $60^{0}$.

| Points | Grading Criteria |
| :---: | :---: |
| 20 | A well-reasoned and correctly executed solution to the problem. |
| 15 | The solution is correct and well-reasoned but contains an arithmetic error or is insufficiently justified. |
| 7 | The solution is correctly started, some intermediate results are obtained, but the further solution is incorrect or missing. |
| 0 | The solution does not meet the above requirements. |