MathNet 1999 — Part 01 Question 25

Let $ABC$ be a triangle with $\angle C = 60^{\circ}$ and $|AC| \lt |BC|$. The point $D$ lies on the side $BC$ and satisfies $|BD| = |AC|$. The side $AC$ is extended to the point $E$ where $|AC| = |CE|$. Prove that $|AB| = |DE|$.