Question

In figure, the side QR of $\Delta$PQR is produced to a point S. If the bisectors of $\angle$PQR and $\angle$PRS meet at point T, then $\angle$QTR$=\frac{1}{2}\angle$QPR.

• A. True
• B. False

Answer 1

The correct option is A True

Given,
Bisectors of ∠PQR & ∠PRS meet at point T.
To prove,
∠QTR = $\frac{1}{2}$∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR
(Exterior angle of a triangle equals to the sum of the two interior angles.)
∠QTR=∠TRS–∠TQR$-\left(1\right)$
$⟹$∠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)
$\frac{1}{2}$∠QPR = ∠TRS – ∠TQR$-\left(2\right)$
Equating (1) and (2)
∠QTR= $\frac{1}{2}$∠QPR