Question

In Fig. 6.44, the side QR of triangle PQR is produced to
a point S. If the bisectors of angle POR and
angle PRS meet at point T, then prove that
QTR= 1/2 QPR.​

Answer 1

Given:

In ΔPQR,

QR is produced to S

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

To prove:

∠QTR = $\frac{1}{2}$ ∠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 = $\frac{1}{2}$ ∠QPR ....... (ii)

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

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

Hence proved

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Answer 2

Answer:

Given, Bisectors of ∠PQRand ∠PRS meet at point T.

To prove: ∠QTR=

2

1

∠QPR.

Proof,

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

⇒∠QTR=∠TRS−∠TQR --- (i)

Also ∠SRP=∠QPR+∠PQR

2∠TRS=∠QPR+2∠TQR

∠QPR=2∠TRS−2∠TQR

⇒

2

1

∠QPR=∠TRS−∠TQR --- (ii)

Equating (i) and (ii),

∴∠QTR=

2

1

∠QPR [henceproved]