In the given diagram, we need to prove that $\triangle ADE$ is similar to $\triangle ABC$ by showing that two pairs of angles are congruent.

Step-by-step reasoning:

1. Angle $\angle DAE$ is a common angle between both triangles $\triangle ADE$ and $\triangle ABC$.

   • So, $\angle DAE = \angle BAC$.

2. Since $DE \parallel BC$ and $AB$ is a transversal line, corresponding angles are equal:

   • $\angle ADE = \angle ABC$.

3. By the AA (Angle-Angle) similarity criterion, if two corresponding angles are equal, the triangles are similar.

   • Thus, $\triangle ADE \sim \triangle ABC$.

Drag and Drop Answer:

• $\angle DAE = \angle BAC$ (common angle)
• $\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$)
• $\triangle ADE \sim \triangle ABC$ (Two respective angles are equal)

Would you like further details on the reasoning or any clarifications?

Here are some related questions for practice:

1. How do you prove triangles are similar using the AA criterion?
2. What other criteria besides AA can be used to prove triangle similarity?
3. If two triangles are similar, what can be said about their corresponding sides?
4. What are alternate interior angles, and how do they help in proving similarity?
5. Can the similarity of triangles be used to find unknown side lengths?

Tip: In geometry, parallel lines often help identify congruent corresponding or alternate angles.