In trapezium ABCE, AE||BC with E = 70°. (1) Calculate the value of x. (2) Show that ABCD is a parallelogram. (3) Show that AB = CE.

Here are the solutions to the geometry problems:

Problem (3): ABCD is a parallelogram

• In a parallelogram, opposite angles are equal. $$\angle DAB = \angle BCD$$ $$80^\circ = 140^\circ - 2y$$ $$2y = 140^\circ - 80^\circ$$ $$2y = 60^\circ$$ $$y = 30^\circ$$
• Since $AD \parallel BC$, alternate interior angles are equal: $\angle ADB = \angle DBC$. From the diagram, $\angle DBC = y$, so $\angle ADB = y = 30^\circ$.
• Since $AB \parallel DC$, alternate interior angles are equal: $\angle ABD = \angle CDB$. From the diagram, $\angle CDB = x$, so $\angle ABD = x$.
• In a parallelogram, consecutive angles are supplementary: $\angle DAB + \angle ADC = 180^\circ$. $$\angle ADC = \angle ADB + \angle CDB = y + x = 30^\circ + x$$ $$80^\circ + (30^\circ + x) = 180^\circ$$ $$110^\circ + x = 180^\circ$$ $$x = 180^\circ - 110^\circ$$ $$x = 70^\circ$$
• Therefore, $x = 70^\circ$ and $y = 30^\circ$.

Problem (4): ABCD is a rhombus

• In a rhombus, all sides are equal, and the diagonals bisect the angles.
• Since $AB = BC$, $\triangle ABC$ is an isosceles triangle.
• Given $\angle ABC = 50^\circ$.
• The base angles of $\triangle ABC$ are equal: $\angle BAC = \angle BCA$.
• The sum of angles in $\triangle ABC$ is $180^\circ$: $$\angle ABC + \angle BAC + \angle BCA = 180^\circ$$ $$50^\circ + \angle BAC + \angle BAC = 180^\circ$$ $$2 \angle BAC = 180^\circ - 50^\circ$$ $$2 \angle BAC = 130^\circ$$ $$\angle BAC = 65^\circ$$
• Since the diagonals of a rhombus bisect the angles, $\angle BAC = \angle CAD = y$ and $\angle BCA = \angle ACD = x$.
• Therefore, $x = 65^\circ$ and $y = 65^\circ$.

Problem (5): Triangle DFG

• The sum of angles in a triangle is $180^\circ$.
• Given $\angle D = 6x - 15^\circ$, $\angle F = 2x + 12^\circ$, $\angle G = 3x$. $$(6x - 15^\circ) + (2x + 12^\circ) + 3x = 180^\circ$$ $$11x - 3^\circ = 180^\circ$$ $$11x = 183^\circ$$ $$x = \frac{183}{11}^\circ$$
• The value of $x$ is $\frac{183}{11}^\circ$.

Problem (6): Triangle ADE with parallel lines

• The markings indicate that $BC = CD = DE$ and $AF = FG = GE$.
• By the converse of the intercept theorem (or Thales's theorem), if segments on two transversals are proportional, then the lines connecting corresponding points are parallel. Thus, $BF \parallel CG \parallel DE$.
• This implies that $\triangle ABF \sim \triangle ACG \sim \triangle ADE$.
• Therefore, corresponding angles are equal: $\angle AFB = \angle AGC = \angle AED$ $\angle ABF = \angle ACG = \angle ADE$
• From the diagram, the angles are given as: $\angle AFB = 5y$ $\angle ADE = 2y$ $\angle AED = \text{angle labeled '2'}$
• From the similarity, $\angle AFB = \angle ADE$ is not necessarily true. It's $\angle AFB = \angle AED$ and $\angle ABF = \angle ADE$.
• So, we have $\angle AFB = 5y$ and $\angle AED = \text{angle labeled '2'}$. Thus, $5y = \text{angle labeled '2'}$.
• We also have $\angle ADE = 2y$.
• In $\triangle ADE$, the sum of angles is $180^\circ$: $$\angle A + \angle ADE + \angle AED = 180^\circ$$ $$\angle A + 2y + 5y = 180^\circ$$ $$\angle A + 7y = 180^\circ$$
• This equation alone cannot solve for $y$ or $\angle A$. There might be missing information or a misinterpretation of the diagram's angle labels. Assuming the labels $5y$ and $2y$ refer to $\angle AED$ and $\angle ADE$ respectively, and the question is asking for $y$.
• If $\angle AED = 5y$ and $\angle ADE = 2y$, then $\angle A = 180^\circ - 7y$.
• However, the diagram shows $5y$ at $F$ and $2y$ at $D$.
• Let's assume the angles are $\angle AFG = 5y$ and $\angle ADE = 2y$.
• If $BF \parallel CG \parallel DE$, then $\angle AFG = \angle ADE$ (corresponding angles). $$5y = 2y$$ $$3y = 0$$ $$y = 0$$
• This result is impossible for an angle in a triangle. There is an inconsistency in the problem statement or diagram for question (6). Without further clarification or a different interpretation, a valid solution for $y$ cannot be found.

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Let's re-evaluate problem (6) with a common alternative interpretation for such diagrams:
Often, the angles $5y$ and $2y$ are given as angles within the parallel lines, for example, $\angle AFG = 5y$ and $\angle GDE = 2y$.
If $BF \parallel CG \parallel DE$, then $\angle AFG = \angle ADE$ (corresponding angles). So $\angle ADE = 5y$.
The angle $2y$ is $\angle GDE$. This means $\angle ADE = \angle ADG + \angle GDE$. This is not consistent with $G$ being on $AD$.

Let's assume the angles are:
$\angle AFG = 5y$
$\angle ADE = 2y$
$\angle AED = \text{angle labeled '2'}$

If $BF \parallel CG \parallel DE$, then $\angle AFG = \angle ADE$ (corresponding angles).
So $5y = 2y$, which means $y=0$. This is impossible.

Given the ambiguity, I cannot provide a definitive numerical answer for problem (6) without making an assumption that contradicts the diagram or leads to an impossible result. I will state the properties that would be used.

Problem (6): Triangle ADE with parallel lines

• The markings indicate $BC = CD = DE$ and $AF = FG = GE$.
• This implies $BF \parallel CG \parallel DE$ by the converse of the intercept theorem.
• Therefore, $\triangle ABF \sim \triangle ACG \sim \triangle ADE$.
• This means corresponding angles are equal: $\angle AFB = \angle AGC = \angle AED$ and $\angle ABF = \angle ACG = \angle ADE$.
• The given angle labels $5y$ and $2y$ are inconsistent with these properties if they refer to $\angle AFG$ and $\angle ADE$ respectively, as it would lead to $y=0$. Without further clarification on which specific angles $5y$ and $2y$ represent, a numerical solution for $y$ cannot be determined.

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Final Answers:
3. ABCD is a parallelogram

• $x = \boxed{\mathbf{70^\circ}}$
• $y = \boxed{\mathbf{30^\circ}}$

4. ABCD is a rhombus

• $x = \boxed{\mathbf{65^\circ}}$
• $y = \boxed{\mathbf{65^\circ}}$

5. Triangle DFG

• $x = \boxed{\mathbf{\frac{183}{11}^\circ}}$

6. Triangle ADE with parallel lines

• Based on the markings, $BF \parallel CG \parallel DE$. This implies $\triangle ABF \sim \triangle ACG \sim \triangle ADE$.
• The angle labels $5y$ and $2y$ as shown in the diagram lead to an inconsistency ($y=0$) when applying properties of parallel lines and similar triangles. Therefore, a numerical value for $y$ cannot be determined from the given information.