4. Point B is internal to segment AC. On one side of line AC, equilateral triangles ABM and BCN are constructed. Lines AN and CM intersect at point L. Calculate the angle CLN.
Solution: Obviously $\triangle \mathrm{ABN} \cong \triangle \mathrm{MBC}$ according to the SAS criterion, (since $\overline{\mathrm{AB}}=\overline{\mathrm{MB}}, \overline{\mathrm{BN}}=\overline{\mathrm{BC}}$ and $\angle \mathrm{ABN}=120^{\circ}=$ $\angle \mathrm{MBC}$ ). Therefore, it follows that $\angle \mathrm{ANB}=$ $\angle \mathrm{MCB}=\alpha$. Since $\angle \mathrm{BKC}=\angle \mathrm{LKN}$ as

![](https://cdn.mathpix.com/cropped/2024_06_05_a08378ec0cd0d184c22fg-2.jpg?height=310&width=390&top_left_y=821&top_left_x=1074)
alternate interior angles, it follows that the third angles in triangles BCK and LNK are also equal, and these are $\angle \mathrm{CBK}=60^{\circ}=\angle \mathrm{NLK}$, hence $\angle \mathrm{CLN}=60^{\circ}$.