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;

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.

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, $\angle A = \angle BCE = 70^\circ$.

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.

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, $\angle B = \angle AEC = 110^\circ$.

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).

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

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)