In trapezoid ABCD, AD || BC, with A = 70^ and C = DC. In figure below, ABFD is trapezum, AC || DG. BEF = 45^ and BFG = 105^. Calculate with reasons the size of the following;

Step 1: Answer question 3.1.1. In trapezium ABCD, AD || BC, with $\angle A = \angle D = 70^\circ$ and EC = DC. 3.1.1. $\angle CED = 70^\circ$ Since AD || BC, and EC is a transversal, $\angle CED$ and $\angle ADE$ are alternate interior angles if AE || DC. However, the diagram shows AD || BC. Given $\angle D = 70^\circ$. Since EC = DC, $\triangle EDC$ is an isosceles triangle. Therefore, $\angle CED = \angle D = 70^\circ$ (angles opposite equal sides). Reason: Angles opposite equal sides.

Step 2: Answer question 3.1.2. 3.1.2. $\angle BCE = 70^\circ$ Since AD || BC, $\angle A$ and $\angle B$ are co-interior angles, so $\angle A + \angle B = 180^\circ$. $\angle A = 70^\circ$, so $\angle B = 110^\circ$. Also, $\angle D = 70^\circ$. In $\triangle EDC$, $\angle ECD = 180^\circ - (\angle CED + \angle D) = 180^\circ - (70^\circ + 70^\circ) = 180^\circ - 140^\circ = 40^\circ$. Since AD || BC, $\angle ADC$ and $\angle BCD$ are co-interior angles. $\angle ADC = \angle ADE + \angle EDC = 70^\circ + 70^\circ = 140^\circ$. This is incorrect. $\angle D$ refers to $\angle ADC$. Given $\angle A = 70^\circ$ and $\angle D = 70^\circ$. Since AD || BC, $\angle D + \angle C = 180^\circ$ (co-interior angles). So, $\angle C = 180^\circ - 70^\circ = 110^\circ$. $\angle C$ is $\angle BCD$. We know $\angle ECD = 40^\circ$. $\angle BCE = \angle BCD - \angle ECD = 110^\circ - 40^\circ = 70^\circ$. Reason: Co-interior angles AD || BC and angles in a triangle. Therefore, $\angle A = \angle BCE = 70^\circ$.

Step 3: Answer question 3.1.3. 3.1.3. $\angle AEC = 110^\circ$ In trapezium ABCD, AD || BC. Consider the line AE as a transversal. $\angle A = 70^\circ$. $\angle AEC$ and $\angle EAD$ are co-interior angles if AB || EC. This is not given. However, $\angle AEC$ is an exterior angle to $\triangle EDC$ if we extend DE. Alternatively, consider quadrilateral ABCE. We have $\angle A = 70^\circ$, $\angle B = 110^\circ$ (from 3.1.4), $\angle BCE = 70^\circ$. The sum of angles in a quadrilateral is $360^\circ$. $\angle AEC = 360^\circ - (\angle A + \angle B + \angle BCE) = 360^\circ - (70^\circ + 110^\circ + 70^\circ) = 360^\circ - 250^\circ = 110^\circ$. Reason: Sum of angles in a quadrilateral.

Step 4: Answer question 3.1.4. 3.1.4. $\angle B = 110^\circ$ Since AD || BC, $\angle A$ and $\angle B$ are co-interior angles. $\angle A + \angle B = 180^\circ$. $70^\circ + \angle B = 180^\circ$. $\angle B = 180^\circ - 70^\circ = 110^\circ$. Reason: Co-interior angles, AD || BC. Therefore, $\angle B = \angle AEC = 110^\circ$.

Step 5: Answer question 3.1.5. 3.1.5. ABCE is parallelogram For ABCE to be a parallelogram, both pairs of opposite sides must be parallel (AB || EC and AE || BC) or one pair of opposite sides parallel and equal (e.g., AE || BC and AE = BC). We are given AD || BC. This means AE || BC. From the calculations: $\angle A = 70^\circ$ $\angle B = 110^\circ$ $\angle BCE = 70^\circ$ $\angle AEC = 110^\circ$ Since $\angle A = \angle BCE = 70^\circ$ and $\angle B = \angle AEC = 110^\circ$, opposite angles are equal. Also, since AE || BC (as AD || BC), and $\angle A + \angle B = 180^\circ$, $\angle A + \angle AEC = 70^\circ + 110^\circ = 180^\circ$. This means AB || EC. Since both pairs of opposite sides are parallel (AE || BC and AB || EC), ABCE is a parallelogram. Reason: Both pairs of opposite sides are parallel (or Opposite angles are equal).

Step 6: Answer question 3.2. 3.2. Complete the following statement: Sum of the interior angles of the triangles are $\boxed{180^\circ}$.

Step 7: Answer question 3.3. In figure below, ABFD is trapezium, AC || DG. $\angle BEF = 45^\circ$ and $\angle BFG = 105^\circ$. Calculate with reasons the size of the following: We need to calculate the size of angles labeled 1, 2, 3. Since AC || DG, we have parallel lines. Angle 1 (at E): $\angle ABE$ and $\angle BEF$ are alternate interior angles if AB || EF. This is not given. However, $\angle ABE$ is not labeled as 1. The angle labeled 1 is $\angle DEB$. Angle 1 (at D): $\angle ADE$. Angle 1 (at F): $\angle BFC$. Let's assume the question asks for the angles labeled with numbers in the diagram.

Angle 1 (at E): This is $\angle DEB$. Since AC || DG, and BE is a transversal, $\angle CBE$ and $\angle BEF$ are alternate interior angles. $\angle BEF = 45^\circ$. So $\angle CBE = 45^\circ$. The angle labeled 1 at E is $\angle DEB$. This is not directly related to $\angle BEF$ by parallel lines. Let's re-examine the diagram. The arrows indicate AC || DG. The angle labeled '1' inside $\triangle ABE$ is $\angle BAE$. The angle labeled '1' inside $\triangle BFG$ is $\angle FBG$. The angle labeled '1' at E is $\angle DEB$. The angle labeled '1' at F is $\angle BFC$.

Let's assume the question asks for the angles labeled with numbers in the diagram. Angle 1 (at E): $\angle BEF = 45^\circ$. This is an angle on a straight line DG. The angle labeled '1' at E is $\angle DEB$. Since AC || DG, $\angle CAB$ and $\angle ABD$ are alternate interior angles. $\angle ABE$ is labeled as '1'. $\angle BAE$ is labeled as '1'. $\angle FBG$ is labeled as '1'. $\angle DEB$ is labeled as '1'. $\angle BFC$ is labeled as '1'.

Let's assume the question asks for the angles labeled with numbers in the diagram, specifically the angles within the triangles. Angle 1 (at A): $\angle BAE$. Angle 2 (at B): $\angle ABE$. Angle 3 (at B): $\angle EBC$. Angle 1 (at E): $\angle DEB$. Angle 2 (at E): $\angle BEF$. Given as $45^\circ$. Angle 1 (at F): $\angle BFC$. Angle 2 (at F): $\angle BFG$. Given as $105^\circ$.

Let's calculate the angles based on the parallel lines AC || DG. Angle 1 (at E, $\angle DEB$): $\angle BEF = 45^\circ$. $\angle DEB$ and $\angle BEF$ are angles on a straight line DG. So, $\angle DEB + \angle BEF = 180^\circ$. $\angle DEB + 45^\circ = 180^\circ$. $\angle DEB = 180^\circ - 45^\circ = 135^\circ$. Reason: Angles on a straight line. So, $\angle 1$ (at E) = $\boxed{135^\circ}$.

Angle 1 (at F, $\angle BFC$): $\angle BFG = 105^\circ$. $\angle BFC$ and $\angle BFG$ are angles on a straight line DG. So, $\angle BFC + \angle BFG = 180^\circ$. $\angle BFC + 105^\circ = 180^\circ$. $\angle BFC = 180^\circ - 105^\circ = 75^\circ$. Reason: Angles on a straight line. So, $\angle 1$ (at F) = $\boxed{75^\circ}$.

Angle 2 (at B, $\angle EBF$): In $\triangle BEF$, we have $\angle BEF = 45^\circ$. Since AC || DG, $\angle EBF$ is not directly related to $\angle BEF$ or $\angle BFG$. However, $\angle EBF$ is an angle in $\triangle BEF$. We need another angle in $\triangle BEF$ or information about the lines. The diagram shows ABFD is a trapezium, which means AB || DF. If AB || DF, then $\angle ABE$ and $\angle BEF$ are alternate interior angles. So, $\angle ABE = \angle BEF = 45^\circ$. The angle labeled '2' at B is $\angle EBF$. The angle labeled '2' at E is $\angle BEF = 45^\circ$. The angle labeled '2' at F is $\angle BFG = 105^\circ$.

Let's assume the question is asking for the angles labeled 1, 2, 3 in the diagram. Angle 1 (at A): $\angle BAE$. Angle 2 (at B): $\angle EBF$. Angle 3 (at B): $\angle FBC$.

Given AC || DG. $\angle BEF = 45^\circ$. $\angle BFG = 105^\circ$.

Consider $\triangle BEF$. We know $\angle BEF = 45^\circ$. We need to find $\angle EBF$ (labeled 2 at B) and $\angle BFE$. Since AC || DG, $\angle CBF$ and $\angle BFG$ are alternate interior angles. So, $\angle CBF = \angle BFG = 105^\circ$. This is the angle labeled '3' at B. So, $\angle 3$ (at B) = $\boxed{105^\circ}$. Reason: Alternate interior angles, AC || DG.

Now, consider $\triangle BFG$. We have $\angle BFG = 105^\circ$. The angle labeled '1' at F is $\angle BFC$. This is $180^\circ - 105^\circ = 75^\circ$. The angle labeled '1' at B is $\angle FBG$. The angle labeled '2' at B is $\angle EBF$.

Let's re-evaluate the labels. The angle labeled '1' at A is $\angle BAE$. The angle labeled '2' at B is $\angle EBF$. The angle labeled '3' at B is $\angle FBC$. The angle labeled '1' at E is $\angle DEB$. The angle labeled '2' at E is $\angle BEF = 45^\circ$. The angle labeled '1' at F is $\angle BFE$. The angle labeled '2' at F is $\angle BFG = 105^\circ$.

Let's find $\angle BFE$ (labeled '1' at F). $\angle BFE$ and $\angle BFG$ are angles on a straight line. $\angle BFE + \angle BFG = 180^\circ$. $\angle BFE + 105^\circ = 180^\circ$. $\angle BFE = 75^\circ$. Reason: Angles on a straight line. So, $\angle 1$ (at F) = $\boxed{75^\circ}$.

Now, find $\angle EBF$ (labeled '2' at B). In $\triangle BEF$, the sum of angles is $180^\circ$. $\angle EBF + \angle BEF + \angle BFE = 180^\circ$. $\angle EBF + 45^\circ + 75^\circ = 180^\circ$. $\angle EBF + 120^\circ = 180^\circ$. $\angle EBF = 180^\circ - 120^\circ = 60^\circ$. Reason: Sum of angles in a triangle. So, $\angle 2$ (at B) = $\boxed{60^\circ}$.

Now, find $\angle FBC$ (labeled '3' at B). Since AC || DG, $\angle FBC$ and $\angle BFG$ are alternate interior angles. $\angle FBC = \angle BFG = 105^\circ$. Reason: Alternate interior angles, AC || DG. So, $\angle 3$ (at B) = $\boxed{105^\circ}$.

The question asks to calculate the size of "the following". It doesn't specify which angles. I've calculated all angles labeled with numbers in the diagram. Summary of 3.3 angles: $\angle 1$ (at E, $\angle DEB$) = $\boxed{135^\circ}$ (Angles on a straight line) $\angle 1$ (at F, $\angle BFE$) = $\boxed{75^\circ}$ (Angles on a straight line) $\angle 2$ (at B, $\angle EBF$) = $\boxed{60^\circ}$ (Sum of angles in a triangle) $\angle 3$ (at B, $\angle FBC$) = $\boxed{105^\circ}$ (Alternate interior angles, AC || DG)