Question

please give me correct answer with explain​

Answer 1

Answer:

In ΔQTR,

∠ TRS = ∠ TQR + ∠ QTR Exterior angle theorem in a triangle

∠ QTR = ∠ TRS - ∠ TQR ......(I)

Also in ΔQPR,

∠ SRP = ∠ QPR + ∠ PQR

2∠ TRS = ∠ QPR + 2∠ TQR ∠TRS and ∠TQR are the bisectors of ∠SRP and ∠PQR respectively

∠ QPR = 2 ∠ TRS - 2 ∠ TQR

∠ TRS - 2 ∠ TQR = 1/2 ∠ QPR .....(II)

From (I) and (II), we get

∠ QTR = 1/2 ∠ QPR