Question

In fig, the side QR of $△PQR$ is produced to a point ${}^{\prime }{S}^{\prime }$. If the bisectors of $\angle PQR$ & $\angle PRS$ meet at point ${}^{\prime }{T}^{\prime }$, then prove that $\angle QTR=\frac{1}{2}\angle QPR$

Answer 1

Exterior Angle 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.

Solution:

Given,

Bisectors of ∠PQR & ∠PRS meet at point T.

To prove,

∠QTR = 1/2∠QPR.

Proof,

∠TRS = ∠TQR +∠QTR

(Exterior angle of a triangle equals to the sum of the two interior angles.)

⇒∠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

Hence proved.