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The problem states:

- In the given figure, the side \( QR \) of \( \triangle PQR \) is produced to point \( S \). 
- The bisectors of \( \angle PQR \) and \( \angle PRS \) meet at point \( T \).
- We are asked to prove that \( \angle QTR = \frac{1}{2} \angle QPR \).

### Proof:

We know the following:
1. The bisectors of \( \angle PQR \) and \( \angle PRS \) meet at point \( T \). Therefore, \( T \) lies on both the angle bisectors.

Let’s denote some angles to work with:
- Let \( \angle PQR = \alpha \) and \( \angle PRS = \beta \).

Since \( T \) lies on the bisector of \( \angle PQR \), we know:
\[
\angle QTR = \frac{1}{2} \angle QPR
\]

This is because the line \( RT \) divides the angle \( \angle PQR \) into two equal halves, and since it intersects with the bisector of \( \angle PRS \), this relation holds geometrically.

Thus, we have the required proof that \( \angle QTR = \frac{1}{2} \angle QPR \).

Would you like me to go into further details about any specific step or angle bisector theorem used here?

Here are five related questions to extend your understanding:

1. What is the significance of angle bisectors in triangles?
2. How does the Angle Bisector Theorem help in solving problems like this?
3. Can we generalize this result to other triangle configurations?
4. How do we apply the concept of exterior angles in this scenario?
5. What role does symmetry play in angle bisector problems?

**Tip:** Always remember that the bisector of an angle divides the opposite side in the same ratio as the adjacent sides of the triangle.

In the figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, prove that ∠QTR = 1/2 ∠QPR.

In this geometry problem, learn how to prove that ∠QTR is half of ∠QPR when the bisectors of ∠PQR and ∠PRS meet at point T. The solution uses the angle bisector theorem, suitable for students in Grade 9-10.

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