Question

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

Answer 1

EXPLANATION.

Side QR of ΔPQR is produced to a point S.

Bisector of ∠PQR and ∠PRS meet at point T.

As we know that,

TQ is the bisector of ∠PQR.

⇒ ∠PQT = ∠TQR = 1/2∠PQR.

TR is a bisector of ∠PRS.

⇒ ∠PRT = ∠TRS = 1/2∠PRS.

As we know that,

In ΔTQR.

⇒ ∠TRS = ∠QTR + ∠TQR. - - - - - (1).

In ΔPQR.

⇒ ∠PRS = ∠PQR + ∠QPR. - - - - - (2).

We can write equation as,

⇒ ∠TRS = 1/∠PRS.

⇒ ∠TQR = 1/2∠PQR.

⇒ 1/2∠PRS = ∠QTR + 1/2∠PQR.

As we know that,

⇒ ∠PRS = ∠PQR + ∠QPR [ Put the value in equation].

⇒ 1/2(∠PQR + ∠QPR) = ∠QTR + 1/2∠PQR.

⇒ 1/2∠PQR + 1/2∠QPR = ∠QTR + 1/2∠PQR.

⇒ 1/2∠PQR - 1/2∠PQR + 1/2∠QPR = ∠QTR.

⇒ 1/2∠QPR = ∠QTR.

Hence Proved.

Answer 2

Solution :-

We know that the sum of exterior angle are always equal to the sum of both interior angles.

Exterior angle = TRS

Interior angle = TQR and QTR

So,

$\sf \angle TQR + \angle QTR = \angle TRS$

$\sf \angle TRS - \angle TQR = \angle QTR (1)$

Now

$\sf \angle SRP = \angle PQR + \angle QPR$

$\sf \angle SRP = 2\angle TRS$

and

$\sf\angle PQR =2 \angle TQR$

$\sf 2\angle TRS = \angle QPR+2\angle TQR$

$\sf 2\angle TRS-2\angle TQR = \angle QPR$

On dividing by 2

$\sf \dfrac{2\angle TRS}{2} - \dfrac{2 \angle TQR}2 = \dfrac{\angle QPR}{2}$

$\sf \angle TRS - \angle TQR = \dfrac{1}{2} QPR(2)$

Now

Both equation are equal so proved