Question

The side QR of triangle pqr is produced to a point s if the bisector of angle special and angle PRS meet at point t qpr and then prove that angle QtR in is equal to 1/2 by angle qpr​

Answer 1

Answer:

Given:

In ΔPQR,

QR is produced to S

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

To prove:

∠QTR = \frac{1}{2}

2

1

∠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 = \frac{1}{2}

2

1

∠QPR ....... (ii)

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

\boxed{\boxed{\underline{\bold{\angle QTR = \frac{1}{2}\angle QPR }}}}

∠QTR=

2

1

∠QPR

Hence proved

Answer 2

hope it's help you mate..