To prove that $\triangle ACB$ is similar to $\triangle ECD$, we use angle-angle similarity. Let's go through the steps in the diagram.

We are given that $ED \parallel AB$.

1. $\angle ACB = \angle ECD$ because these are vertically opposite angles.
2. $\angle CAB = \angle CED$ because these are alternate angles formed by the transversal $AC$ cutting the parallel lines $AB \parallel ED$.

Thus, by the 2 respective angles equal condition, $\triangle ACB$ is similar to $\triangle ECD$.

Here’s the correct drag-and-drop sequence:

• $\angle ACB = \angle ECD$ → Vertically opposite angles
• $\angle CAB = \angle CED$ → Alternate angles, $ED \parallel AB$
• Conclusion: $\triangle ACB \sim \triangle ECD$ → 2 respective angles equal

Would you like further clarification on these steps?

Related Questions:

1. What is the definition of similar triangles?
2. Why do alternate angles form when a transversal cuts two parallel lines?
3. What other properties can be used to prove triangle similarity?
4. Can two triangles be similar if they do not share any common angles?
5. What are corresponding sides in similar triangles, and how do they relate?

Tip:

When working with parallel lines, look for transversal lines that create corresponding, alternate, or vertically opposite angles to help prove similarity or congruence.