ABCD is a parallelogram. AM bisects A. AB = AM. C = 120^. Calculate the sizes of all interior angles.

In a parallelogram, opposite angles are equal, and consecutive angles are supplementary (add up to $180^\circ$).
Given $\hat{C} = 120^\circ$.

Since opposite angles are equal:

$$\hat{A} = \hat{C}$$ $$\hat{A} = 120^\circ$$

Since consecutive angles are supplementary:

$$\hat{B} + \hat{C} = 180^\circ$$ $$\hat{B} + 120^\circ = 180^\circ$$ $$\hat{B} = 180^\circ - 120^\circ$$ $$\hat{B} = 60^\circ$$

Since opposite angles are equal:

$$\hat{D} = \hat{B}$$ $$\hat{D} = 60^\circ$$

The problem states that AM bisects $\hat{A}$ and $AB = AM$.
If $\hat{A} = 120^\circ$ and AM bisects it, then:

$$\angle BAM = \frac{1}{2} \hat{A} = \frac{1}{2} \times 120^\circ = 60^\circ$$

In $\triangle ABM$, we are given $AB = AM$. This means $\triangle ABM$ is an isosceles triangle. The angles opposite the equal sides are equal:

$$\angle ABM = \angle AMB$$

We know $\angle ABM$ is the same as $\hat{B}$ of the parallelogram, which is $60^\circ$.
So, $\angle AMB = 60^\circ$.
Now, check the sum of angles in $\triangle ABM$:

$$\angle BAM + \angle ABM + \angle AMB = 60^\circ + 60^\circ + 60^\circ = 180^\circ$$

This is consistent, as the sum of angles in a triangle must be $180^\circ$. This also means $\triangle ABM$ is an equilateral triangle. The given information is consistent with the calculated angles of the parallelogram.

The interior angles of the parallelogram are:

$$\hat{A} = \boxed{120^\circ}$$ $$\hat{B} = \boxed{60^\circ}$$ $$\hat{C} = \boxed{120^\circ}$$ $$\hat{D} = \boxed{60^\circ}$$