2. Given a triangle $ABC$, in which $AB = BC$. On the side $BC$, there is a point $D$ such that $CD = AC$. Point $E$ on the ray DA is such that $DE = AC$. Which segment is longer - $EC$ or $AC$? (I. Rubanov)
Answer. Segment $A C$ is not shorter than segment $E C$. Solution. If triangle $A B C$ is equilateral, segments $A C$ and $E C$ are equal. Consider the case when triangle $A B C$ is not equilateral. Since $B C>C D=A C, A C-$ is the smallest side of the isosceles triangle $A B C$, and the angle opposite to it, $\angle A B C$, is the smallest angle of triangle $A B C$. Therefore, $\angle A B C<60^{\circ}$. Then from triangle $A C D$ we have $\angle A D C=\left(180^{\circ}-\angle A C B\right) / 2>60^{\circ}$. Thus, in triangle $C D E$, $\angle D E C>60^{\circ}>\angle E D C=\angle A D C$, from which it follows that $A C=C D>C E$.