2. Point $B$ is internal to segment $\overline{A C}$. In the same half-plane relative to line $A C$, equilateral triangles $\triangle A B M$ and $\triangle B C N$ are constructed. Lines $A N$ and $C M$ intersect at point $L$. Determine the angle $\measuredangle C L N$.
Solution. First, we note that $\measuredangle M B N=60^{\circ}$. This means that $\measuredangle A B N=\measuredangle M B C=120^{\circ}$. From the equilateral nature of triangles $A B M$ and $B C N$, we have that $\overline{A B}=\overline{B M}$ and $\overline{B C}=\overline{B N}$. By the SAS (Side-Angle-Side) congruence criterion, we have $\triangle A B N \cong \triangle M B C$, from which it follows that $\measuredangle A N B=\measuredangle M C B=\alpha$. Let $K$ be the intersection point of lines $M C$ and $B N$. From $\measuredangle B K C=\measuredangle M K N=\beta$ (as

![](https://cdn.mathpix.com/cropped/2024_06_05_3cde5d119ebb42d699c4g-1.jpg?height=298&width=490&top_left_y=1841&top_left_x=990)
vertical angles), in $\triangle B K C$ we have that

$$
\alpha+\beta=180^{\circ}-60^{\circ}=120^{\circ}
$$

Finally, from $\triangle L K N$ we get that

$$
\measuredangle K L N=180^{\circ}-(\alpha+\beta)=60^{\circ}
$$

which means that $\measuredangle C L N=60^{\circ}$.