The problem in the image asks to prove that triangles △QRU and △TSU are similar, given that QR ∥ ST.

Step-by-step proof:

1. Given: QR ∥ ST (QR is parallel to ST)

2. Angles formed by transversals:

   • Since QR is parallel to ST and QS is a transversal, the corresponding angles ∠QRU and ∠TSU are equal (alternate interior angles are equal for parallel lines). Therefore: $\angle QRU = \angle TSU$

   • Similarly, since QR is parallel to ST and RT is a transversal, the corresponding angles ∠QUR and ∠TUS are also equal. Hence: $\angle QUR = \angle TUS$

3. Third angle equality:

   • In triangles, if two angles are equal, the third angle must also be equal by the Angle Sum Property. Thus: $\angle Q = \angle T$

4. Conclusion:

   • Since all corresponding angles in △QRU and △TSU are equal, by the Angle-Angle (AA) similarity criterion, the triangles △QRU and △TSU are similar: $\triangle QRU \sim \triangle TSU$

Final Answer:

We have proven that △QRU ∼ △TSU using the AA similarity criterion.

Would you like more details or have any specific questions?

Here are some related questions you might consider:

1. How does the transversal rule help in proving triangle similarity?
2. What is the Angle-Angle (AA) criterion for triangle similarity?
3. Can triangles with different orientations be similar?
4. How would the solution change if QR was not parallel to ST?
5. Can similarity be used to determine the ratio of corresponding sides?

Tip: When proving triangle similarity, always check for parallel lines or common angles, as they often provide key information for the proof.