# In Fig. 6.44, the side QR of △PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR

**Solution:**

Given: TQ and TR are the bisectors of ∠PQR and ∠PRS respectively

To Prove: ∠QTR = 1/2 ∠QPR

According to the exterior angle theorem of a triangle, if a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

∠PRS = 2∠TRS.......(i) [Since TR is the angle bisector of ∠PRS]

∠PQR = 2∠TQR .........(ii) [Since TQ is the angle bisector of ∠PQR]

Now, in △TQR

∠TRS = ∠TQR +∠QTR [Exterior angle theorem of a triangle]

∠QTR = ∠TRS - ∠TQR.......... (iii)

Similarly, in △PQR

∠PRS = ∠QPR + ∠PQR [Exterior angle theorem of a triangle]

2∠TRS = ∠QPR + 2∠TQR [From (i) and (ii)]

∠QPR = 2∠TRS - 2∠TQR

∠QPR = 2(∠TRS - ∠TQR)

∠QPR = 2∠QTR [From (iii)]

∠QTR = 1/2 ∠QPR

Hence proved, ∠QTR = 1/2 ∠QPR.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 6

**Video Solution:**

## In Fig. 6.44, the side QR of △PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.

NCERT Maths Solutions Class 9 Chapter 6 Exercise 6.3 Question 6

**Summary:**

In Fig. 6.44, if the side QR of △PQR is produced to a point S and the bisectors of ∠PQR and ∠PRS meet at point T, then it is proved that ∠QTR = 1/2 ∠QPR.

**☛ Related Questions:**

- In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∠XYZ, find ∠OZY and ∠YOZ.
- In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
- In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
- In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65° then find the values of x and y.

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