In the figure below A = x, ACE = 50°, BEC = 130°, and ECG = 2x – 110°. 1.1 Determine, giving reasons, the value of x. 1.2 Is AB parallel to FG? Give reasons for your answer.

Angles on a straight line sum to $180^\circ$. Since A, E, B form a straight line, $\angle AEC$ and $\angle BEC$ are supplementary.

$$\angle AEC + \angle BEC = 180^\circ$$

Given $\angle BEC = 130^\circ$:

$$\angle AEC + 130^\circ = 180^\circ$$ $$\angle AEC = 180^\circ - 130^\circ = 50^\circ$$

Reason: Angles on a straight line.

The sum of angles in a triangle is $180^\circ$.

$$\angle A + \angle ACE + \angle AEC = 180^\circ$$

Given $\angle A = x$ and $\angle ACE = 50^\circ$. From Step 1, $\angle AEC = 50^\circ$.

$$x + 50^\circ + 50^\circ = 180^\circ$$ $$x + 100^\circ = 180^\circ$$ $$x = 180^\circ - 100^\circ$$ $$x = 80^\circ$$

Reason: Sum of angles in a triangle.

The final answer is $\boxed{\boldsymbol{80^\circ}}$.

1.1.2

Given $\angle ECG = 2x - 110^\circ$. Substitute $x = 80^\circ$:

$$\angle ECG = 2(80^\circ) - 110^\circ$$ $$\angle ECG = 160^\circ - 110^\circ$$ $$\angle ECG = 50^\circ$$

From 1.1.1, we found $\angle AEC = 50^\circ$.
We just calculated $\angle ECG = 50^\circ$.
Since $\angle AEC$ and $\angle ECG$ are alternate interior angles formed by the transversal line CD intersecting lines AB and FG, and they are equal:

$$\angle AEC = \angle ECG = 50^\circ$$

Therefore, AB is parallel to FG.

Alternatively, using co-interior angles:
We have $\angle BEC = 130^\circ$ and $\angle ECG = 50^\circ$.
These are co-interior angles formed by the transversal line CD intersecting lines AB and FG.

$$\angle BEC + \angle ECG = 130^\circ + 50^\circ = 180^\circ$$

Since the co-interior angles are supplementary, AB is parallel to FG.

The final answer is $\boxed{\text{Yes, AB is parallel to FG because the alternate interior angles (\angle AEC and \angle ECG) are equal, or because the co-interior angles (\angle BEC and \angle ECG) are supplementary.}}$.