Question

, the side QR of ∆ PQR is produced to a point S. If the bisectors of <PQR and <PRS meet at point T, then prove that <QTR =½<QPR.​

Answer 1

GIVEN :-

• The sides QR of ∆ PQR is produced to a point s.
• Bisectors of ∠PQR and ∠PRS meet at point T.

TO PROVE :-

• ∠QTR = 1/2 ∠QPR.

PROOF :-

✭ In ∆ PQR By exterior angle property,

➟ ∠QPR + ∠PQR = ∠PRS

★ Here ∠PQR = ∠PQT + ∠TQR

➟ ∠QPR + ∠PQT + ∠TQR = ∠PRS

★ ∠PRS = ∠PRT + ∠TRS

➟ ∠QPR + ∠PQT + ∠TQR = ∠PRT + ∠TRS

★ Here ∠PQT =∠TQR and ∠PRT =∠TRS

➟ ∠QPR + 2∠TQR = 2∠TRS

➟ 1/2(∠QPR + 2∠TQR) = ∠TRS. .....Eq(1)

✭ In ∆ TQR By exterior angle property,

➙ ∠TQR + ∠QTR = ∠TRS. .....Eq(2)

Here the R.H.S of equation 1 and 2 are equal so the L.H.S of both Equation 1 and 2 will be also equal.

➙ 1/2(∠QPR + 2∠TQR) = ∠TQR + ∠QTR

➙ ∠QPR + 2∠TQR = 2(∠TQR + ∠QTR)

➙ ∠QPR + 2∠TQR = 2∠TQR + 2∠QTR

➙ ∠QPR = 2∠TQR + 2∠QTR - 2∠TQR

➙ ∠QPR = 2∠QTR

➙ ½∠QPR = ∠QTR

Hence proved.

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• In the attachment figure it is given that,

1. The side QR of ∆PQR is produced to a point S .
2. The bisector of ⟨PQR & ⟨PRS meet at point T .

$\huge\bf{\underline{\underline{\gray{TO\:PROVE:-}}}}$

• ⟨QTR = 1/2 ⟨QPR

$\huge\bf{\underline{\underline{\gray{PROOF:-}}}}$

✞︎ It is given that,

• TQ is the bisector of ⟨PQR .

➪ ⟨PQT = ⟨TQR = 1/2 ⟨PQR

✞︎ Also it is given that,

• TR is the bisector of ⟨PRS .

➪ ⟨PRT = ⟨TRS = 1/2 ⟨PRS

☯︎ Now, in ∆PQR

• ⟨PRS is the external angle .

⇒ ⟨PRS = ⟨QPR + ⟨PQR -----(1)

☯︎ Again, in ∆TQR

• ⟨TRS is the external angle .

⇒ ⟨TRS = ⟨TQR + ⟨QTR ------(2)

☞ Putting "⟨TRS = 1/2 ⟨PRS" & "⟨TQR = 1/2 ⟨PQR" .

➝ 1/2 ⟨PRS = 1/2 ⟨PQR + ⟨QTR

☞ Putting "⟨PRS = ⟨QPR + ⟨PQR", from equation (1)

➟ 1/2 [⟨QPR + ⟨PQR] = 1/2 ⟨PQR + ⟨QTR

➟ 1/2 ⟨QPR + 1/2 ⟨PQR = 1/2 ⟨PQR + ⟨QTR

➟ 1/2 ⟨QPR = ⟨QTR

➟ ⟨QTR = 1/2 ⟨QPR

✔️ Hence, Proved .