Question

Side QR of ∆PQR is produced to a point S as shown in the figure.The bisector

of ˂P meets QR at T. Prove that ˂PQR+ ˂PRS=2˂PTR.​

Answer 1

Answer:

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Step-by-step explanation:

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

Proved.

Step-by-step explanation:

Given :

• ∠PQT=∠TQR
• ∠PRT=∠TRS

To Prove :

$\bf \angle QTR=\dfrac{1}{2}(\angle QPR)$

We know that :

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

So,

⇒ ∠PRS=∠QPR + ∠PQR

⇒∠QPR=∠PRS - ∠PQR

⇒ ∠QPR=2∠TRS - 2∠TQR

⇒ ∠QPR=2(∠TRS - ∠TQR)

⇒ 2(∠TQR+∠QTR - ∠TQR)   [ ∠TRS=∠TQR +∠QTR ]

We also know :

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

$\implies \sf \angle QPR=2\;(\angle QTR)\\ \\ \\\implies \sf \angle QTR=\dfrac{1}{2}(\angle QPR)$

Hence Proved.