First, model $\frac{1}{3}$ using a grid.

$\frac{1}{3}$

Add a vertical line down the middle of the model so that it is divided into six equal parts, but the same area of the whole is shaded.

$\frac{1}{3}$

The second model is divided into six equal parts, and two parts are shaded. So, the second model shows $\frac{2}{6}$.
Since the model of $\frac{2}{6}$ has the same area of the whole shaded as the model of $\frac{1}{3}$, $\frac{2}{6}$ is equivalent to $\frac{1}{3}$.

Next, add two more vertical lines to the model so that it is divided into twelve equal parts, but the same area of the whole is shaded.

$\frac{1}{3}$		$\frac{2}{6}$

The third model is divided into twelve equal parts, and four parts are shaded. So, the third model shows $\frac{4}{12}$.
Since the model of $\frac{4}{12}$ has the same area of the whole shaded as the models of $\frac{1}{3}$ and $\frac{2}{6}$, $\frac{4}{12}$ is equivalent to $\frac{2}{6}$ and $\frac{1}{3}$.