Question

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)

Answer 1

Hello mate ☺

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Solution:

∠PQT=∠TQR               (Given)

∠PRT=∠TRS               (Given)

To Prove:  ∠QTR=1/2(∠QPR)

∠PRS=∠QPR+∠PQR

(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)

⇒∠QPR=∠PRS−∠PQR

⇒∠QPR=2∠TRS−2∠TQR

⇒∠QPR=2(∠TRS−∠TQR)

=2(∠TQR+∠QTR−∠TQR)                          (∠TRS=∠TQR+∠QTR)

(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)

⇒∠QPR=2(∠QTR)

⇒∠QTR=1/2(∠QPR)

Hence Proved

I hope, this will help you.☺

Thank you______❤

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Answer 2

Hello mate here is your answer

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Solution:

Solution:Given,

Solution:Given,Bisectors of ∠PQR & ∠PRS meet at point T.

To prove,

To prove,∠QTR = 1/2∠QPR.

Proof,

Proof,∠TRS = ∠TQR +∠QTR

Proof,∠TRS = ∠TQR +∠QTR(Exterior angle of a triangle equals to the sum of the two interior angles.)

⇒∠QTR=∠TRS–∠TQR — (i)

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]⇒∠QPR= 2∠TRS – 2∠TQR

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]⇒∠QPR= 2∠TRS – 2∠TQR⇒∠QPR= 2(∠TRS – ∠TQR)

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]⇒∠QPR= 2∠TRS – 2∠TQR⇒∠QPR= 2(∠TRS – ∠TQR)⇒ 1/2∠QPR = ∠TRS – ∠TQR — (ii)

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]⇒∠QPR= 2∠TRS – 2∠TQR⇒∠QPR= 2(∠TRS – ∠TQR)⇒ 1/2∠QPR = ∠TRS – ∠TQR — (ii)Equating (i) and (ii)

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]⇒∠QPR= 2∠TRS – 2∠TQR⇒∠QPR= 2(∠TRS – ∠TQR)⇒ 1/2∠QPR = ∠TRS – ∠TQR — (ii)Equating (i) and (ii)∠QTR= 1/2∠QPR

⇒∠QTR=∠TRS–∠TQR — (i)∠SRP = ∠QPR + ∠PQR⇒ 2∠TRS = ∠QPR + 2∠TQR[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]⇒∠QPR= 2∠TRS – 2∠TQR⇒∠QPR= 2(∠TRS – ∠TQR)⇒ 1/2∠QPR = ∠TRS – ∠TQR — (ii)Equating (i) and (ii)∠QTR= 1/2∠QPRHence proved.

Hope it helps, ✌️☺️