Question

Side QR of ∆PQR is produced to point S. The bisector of angle P meets QR at T . prove that angle PQR+anglePRS=2 anglePTR.

Answer 1

Hello mate ☺

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Solution:

∠PQT=∠TQR               (Given)

∠PRT=∠TRS               (Given)

To Prove:  ∠QTR=1/2(∠QPR)

∠PRS=∠QPR+∠PQR

(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)

⇒∠QPR=∠PRS−∠PQR

⇒∠QPR=2∠TRS−2∠TQR

⇒∠QPR=2(∠TRS−∠TQR)

=2(∠TQR+∠QTR−∠TQR)                          (∠TRS=∠TQR+∠QTR)

(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)

⇒∠QPR=2(∠QTR)

⇒∠QTR=1/2(∠QPR)

Hence Proved

I hope, this will help you.☺

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Answer 2

Answer:

$\angle QTR = \frac{1}{2} \angle QPR$. It is proved

Step-by-step explanation:

Given

$T Q$ is the bisector of $\angle P Q R .$

So, $\angle \mathrm{PQT}=\angle \mathrm{TQR}=\frac{1}{2} \angle \mathrm{PQR}$

Also,

$\mathrm{TR} is the bisector of \angle \mathrm{PRS}$

So,$\angle \mathrm{PRT}=\angle \mathrm{TRS}=\frac{1}{2} \angle \mathrm{PRS}$

In $\triangle \mathrm{PQR}$,

$\angle \mathrm{PRS}$is the external angle

$\angle \mathrm{PRS}=\angle \mathrm{QPR}+\angle \mathrm{PQR} \quad\left(\begin{array}{c}\text { External angle is sumof two } \\ \text { interior opposite angles }\end{array}\right)$

$\ldots(1)\\ In \triangle \mathrm{TQR} ,$

$\angle TRS$ is the external angle

$\angle \mathrm{TRS}=\angle \mathrm{TQR}+\angle \mathrm{QTR} \quad\left(\begin{array}{c}\text { External angle is sumof two } \\ \text { interior opposite angles }\end{array}\right) \quad \ldots(2)$

Putting $\angle T R S=\frac{1}{2} \angle P R S \& \angle T Q R=\frac{1}{2} \angle P Q R \frac{1}{2} \angle \mathrm{PRS}=\frac{1}{2} \angle \mathrm{PQR}+\angle \mathrm{QTR}$

$\frac{1}{2} \angle \mathrm{PRS}=\frac{1}{2} \angle \mathrm{PQR}+\angle \mathrm{QTR}$

Putting $\angle P R S=\angle Q P R+\angle P Q R from (1)$

$\frac{1}{2}(\angle \mathrm{QPR}+\angle \mathrm{PQR})=\frac{1}{2} \angle \mathrm{PQR}+\angle \mathrm{QTR} \frac{1}{2} \angle \mathrm{QPR}+\frac{1}{2} \angle \mathrm{PQR}=\frac{1}{2} \angle \mathrm{PQR}+\angle \mathrm{QTR} \frac{1}{2} \angle \mathrm{QPR}+\frac{1}{2} \\\angle \mathrm{PQR}-\frac{1}{2} \angle \mathrm{PQR}=\angle \mathrm{QTR}$

$\frac{1}{2} \angle \mathrm{QPR}=\angle \mathrm{QTR}$

$\angle \mathrm{QTR}=\frac{1}{2}\angle \mathrm{QPR}$

Hence it is proved

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