Take B = (0, 0), C = (12, 0) and A = (3, 9)…

Step 1. The point D. $D$ is $\dfrac{2}{3}$ of the way from $A(3, 9)$ to $B(0, 0)$: $D = (3 - 2, 9 - 6) = (1, 3)$.

Step 2. The point E. $E$ is $\dfrac{2}{3}$ of the way from $A(3, 9)$ to $C(12, 0)$: $E = (3 + 6, 9 - 6) = (9, 3)$.

Step 3. Parallel. $D$ and $E$ are both at height $3$, and $BC$ lies along the $x$-axis. So $DE \parallel BC$, as the theorem says.