We are given that $AB \parallel DE$, $\angle ABC = 120^\circ$, and $\angle CDE = 135^\circ$. We are asked to find $\angle BCD$.

Let's break it down step by step:

1. Since $AB \parallel DE$ and $BC$ is a transversal, the corresponding angles theorem tells us that: $\angle ABC + \angle BCD = 180^\circ$ because they form a linear pair.

2. Given $\angle ABC = 120^\circ$, we can substitute this into the equation to find $\angle BCD$: $\angle BCD = 180^\circ - 120^\circ = 60^\circ$

Thus, $\angle BCD = 60^\circ$.

Would you like further details or clarification on this solution?

Related Questions:

1. What is the corresponding angles theorem, and why does it apply here?
2. How do parallel lines affect the angles formed by a transversal?
3. What is the sum of the interior angles of a triangle, and how can it be applied to find unknown angles?
4. If $\angle CDE = 135^\circ$, what can you infer about $\angle CEB$ if $BC$ is extended further?
5. What other properties of parallel lines are useful for solving geometry problems like this?

Tip:

In problems with parallel lines, always look for pairs of corresponding, alternate interior, or co-interior angles—they can help simplify complex figures.