Given that AB = AE, ABE is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal.

Given that $AB = AE$, $\triangle ABE$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal.
We are given $\angle ABE = x$.
Therefore, $\angle AEB = \angle ABE = x$.

The sum of the interior angles in $\triangle ABE$ is $180^\circ$.

Substitute the known angle values:

This proves the first part of the question.

From the diagram, $AB \parallel DC$ (indicated by the arrows on the lines). AD is a transversal line intersecting these parallel lines.
Therefore, the consecutive interior angles $\angle A$ and $\angle D_1$ are supplementary, meaning their sum is $180^\circ$.

Substitute $\angle A = 180^\circ - 2x$ (from Part 1) and $\angle D_1 = 4x - 40^\circ$ (given):

Combine like terms:

Subtract $140^\circ$ from both sides:

Divide by 2: