Question

In the given figure, the side of QR of ∆POR is produced to a point S. If the bisector of angle PQR and angle PRS
meet at point I, then prove that angle QTR = 1/2 angle QPR
​

Answer 1

Answer:

OPEN the pic you will get your ANSWER

Answer 2

Given:

In ΔPQR,

QR is produced to S

The bisectors of ∠PQR & ∠PRS meet at a point T

To prove:

∠QTR = 1/2∠QPR

Solution:

Concept to be used: The exterior angle of a triangle is equal to the sum of two interior opposite angles of the triangle

Considering the exterior angle of ΔTQR, we have

∠TRS = ∠TQR + ∠QTR

⇒ ∠QTR = ∠TRS - ∠TQR ....... (i)

Considering the exterior angle of Δ PQR, we have

∠PRS = ∠PQR + ∠QPR

[∵ QT is the bisector of ∠PQR & TR is the bisector of ∠PRS (as shown in the fig)]

⇒ 2∠TRS = 2∠TQR + ∠QPR

⇒ ∠QPR = 2∠TRS - 2∠TQR

⇒ ∠QPR = 2[∠TRS - ∠TQR]

⇒ ∠TRS - ∠TQR = 1/2∠QPR ....... (ii)

Now, on comparing (i) & (ii), we get

∠QTR= 1/2∠QPR

Hence proved

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