Problem 298

Since $CD=DE=EC$, then $\triangle CDE$ is equilateral, which means that $\angle DEC=60^{\circ}$. Since $\angle DEB$ is a straight angle, then $\angle CEB=180^{\circ}-\angle DEC=180^{\circ}-60^{\circ}=120^{\circ}$. Since $CE=EB$, then $\triangle CEB$ is isosceles with $\angle ECB=\angle EBC$. Since $\angle ECB+\angle CEB+\angle EBC=180^{\circ}$, then $2 \times \angle EBC+120^{\circ}=180^{\circ}$, which means that $2 \times \angle EBC=60^{\circ}$ or $\angle EBC=30^{\circ}$. Therefore, $\angle ABC=\angle EBC=30^{\circ}$.