Question

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Answer 1

Answer:

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.

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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.

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Answer 2

GIVEN:-

• The side of QR of ∆PQR is Produced to a point S.

• if the Bisector of $\rm{\angle{PQR}\:and\:\angle{PRS}}$ meet at the Point T.

TO PROVE

• $\rm{\angle{QTR}=\dfrac{1}{2}\angle{QPR}}$.

PROPERTY USED:-

• Exterior angle Property i.e the Sum of two Opposite interior angle is equal to the external angle.

Now,

In ∆QTR

$\implies\rm{\angle{TQR}+\angle{QTR}=\angle{TRS}}$(Exterior angle Property)

$\implies\rm{\angle{QTR}=\angle{TRS}-\angle{TQR}}$.......1

Now,

In ∆PQR

$\implies\rm{\angle{PQR}+\angle{QPR}=\angle{PRS}}$(Exterior angle Property).

[TR is the Bisector of <PRS and TQ is the Bisector of <PQR.]

• $\rm{\dfrac{1}{2}\angle{PQR}=\angle{TQR}=\angle{PQR}=\angle{2TQR}}$.......2

• Similarly, $\rm{\angle{PRS}=\angle{2TRS}}$..........3

So, From 2 and 3

$\implies\rm{\angle{2TQR}+\angle{QPR}=\angle{2TRS}}$

$\implies\rm{\angle{QPR}=2(\angle{TRS}-\angle{TQR})}$

Substituting the value of eq1

$\implies\rm{\angle{QPR}=2(\angle{QTR})}$

or,

$\implies\rm{\dfrac{1}{2}\angle{QPR}=\angle{QTR}}$