Question

hi. guys


it's urgent
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Answer 1

Answer:

sorry for cut it is right proving

Answer 2

Answer:

∠QTR = (1/2) ∠QPR

Step-by-step explanation:

in Δ PQR

∠PRS = ∠PQR  + ∠QPR  ( External angle of Triangle = Sum of two opposite angles)

in Δ QTR

∠TRS = ∠TQR + ∠QTR

Multiplying by 2

2∠TRS =  2∠TQR + 2∠QTR

RT is bisector of ∠PRS

=> 2∠TRS = ∠PRS

QT is bisector of ∠PQR

=> 2∠TQR = ∠PQR

=> ∠PRS  = ∠PQR + 2∠QTR

    ∠PRS = ∠PQR  + ∠QPR

Equating both

∠PQR + 2∠QTR = ∠PQR  + ∠QPR

=> 2∠QTR =  ∠QPR

=> ∠QTR = (1/2) ∠QPR

QED

Proved