ABCD is a trapezium with AB || BC, AB=AD and C = 180°. Determine the unknown angles.

• ABCD is a trapezium.
• The problem states $AB \parallel BC$, which is a typo. For a trapezium, one pair of opposite sides must be parallel. From the diagram, it's clear that $AB \parallel DC$. We will proceed with this assumption.
• $AB = AD$. This means $\triangle ABD$ is an isosceles triangle.
• $\angle C = 80^\circ$. (The text says $180^\circ$, which is impossible for an angle in a quadrilateral; we use the diagram's $80^\circ$).
• We need to find the unknown angles: $a = \angle CDA$, $b = \angle ABD$, $d = \angle ABC$, $e = \angle DAB$.

Since $AB \parallel DC$, the consecutive interior angles on the transversal $BC$ are supplementary:

Also, since $AB \parallel DC$, the alternate interior angles formed by the transversal $BD$ are equal:

Given $AB = AD$, $\triangle ABD$ is an isosceles triangle. The base angles opposite the equal sides are equal:

From Step 2 and Step 3, we have $\angle ADB = b$ and $\angle BDC = b$.
Therefore, $\angle CDA$ is the sum of these two angles:

In $\triangle ABD$, the sum of angles is $180^\circ$:

In a trapezium where the non-parallel sides are equal, it is an isosceles trapezium. In this case, the condition $AB=AD$ along with $AB \parallel DC$ and $\angle C = 80^\circ$ implies that $AD=BC$, making it an isosceles trapezium.
In an isosceles trapezium, the base angles are equal:

From Step 4, we have $\angle a = 2b$.
Substitute $\angle a = 80^\circ$:

From Step 5, we have $\angle e + 2b = 180^\circ$.
Substitute $2b = 80^\circ$:

Summary of unknown angles:

• $\angle a = \angle CDA = \boxed{\mathbf{80^\circ}}$
• $\angle b = \angle ABD = \boxed{\mathbf{40^\circ}}$
• $\angle d = \angle ABC = \boxed{\mathbf{100^\circ}}$
• $\angle e = \angle DAB = \boxed{\mathbf{100^\circ}}$