Use the diagram below to prove that the opposite sides of a parallelogram are equal.

Given that PQRS is a parallelogram, we know that its opposite sides are parallel: $PQ \parallel SR$ and $PS \parallel QR$.

Since $PQ \parallel SR$ and PR is a transversal, the alternate interior angles are equal:

Since $PS \parallel QR$ and PR is a transversal, the alternate interior angles are equal:

The diagonal PR is common to both $\triangle PQR$ and $\triangle RSP$.

By the Angle-Side-Angle (ASA) congruence criterion, $\triangle PQR$ is congruent to $\triangle RSP$:

Since the triangles are congruent, their corresponding sides are equal:

Therefore, the opposite sides of a parallelogram are equal.

1.2 In the diagram below ABCD is a parallelogram with $\hat{G} = 9x + 12^\circ$ and $\hat{F} = 15x - 24^\circ$.
(Assuming the parallelogram in the diagram is EFGH, and the given angles $\hat{G}$ and $\hat{F}$ refer to $\angle G$ and $\angle F$ of parallelogram EFGH.)

1.2.1 Calculate the value of x.

In a parallelogram, consecutive angles are supplementary, meaning they add up to $180^\circ$.
Therefore, $\hat{G} + \hat{F} = 180^\circ$.

Substitute the given expressions for $\hat{G}$ and $\hat{F}$:

Combine like terms:

Add $12^\circ$ to both sides:

Divide by 24:

The value of x is $\boxed{\text{8}}$.

1.2.2 Determine the size of $\hat{E}$.

In a parallelogram, opposite angles are equal.
Therefore, $\hat{E} = \hat{G}$.

Substitute $x = 8$ into the expression for $\hat{G}$:

Since $\hat{E} = \hat{G}$:

The size of $\hat{E}$ is $\boxed{\text{84}^\circ}$.