Question

In fig side QR of traingle PQR is produced to S point .if the bisector of angle PQR and Angle PRS meet at point T then prove that Angle QTR = 1/2 QPR

Answer 1

$\huge{\bf{\underline{\star{\red{Solution :}}}}}$

Given,

• ∠PQR and ∠PRS of bisector meet at S point

To Proof :

• ∠QTR = 1/2QPR

Proof :

$\implies \bold{\angle TQR = \frac{1}{2} \angle PQR}$..........1) Equation

$\implies \bold{\angle TRQ = \angle R + \frac{1}{2} \angle PRS}$

∵

$\implies \bold{ \angle PRS = \angle P + \angle Q}$

$\implies \bold{\angle TRQ = \angle R + [ \frac{\angle P + \angle Q}{2}} ]$

$\implies \bold{\frac{2\angle R + \angle P + \angle Q}{2}}$

$\implies \bold{\frac{\angle R + \angle R + \angle P + \angle Q}{2}}$

$\implies \bold{ \angle TRQ = \frac{\angle R + 180^{\circ}}{2}}$........................2) Equation

$\rule{200}2$

$\implies \bold{\angle TQR + \angle TRQ + \angle QTR = 180^{\circ}}$

$\implies \bold{\frac{\angle Q}{2} + \frac{\angle R + 180^{\circ}}{2} + 2\angle QTR = 180^{\circ} \times 2 }$

$\implies \bold{ \angle Q + \angle R + 180^{\circ} + 2\angle QTR = 360^{\circ}}$

$\implies \bold{2\angle QTR = 360^{\circ} - 180{\circ} - \angle Q - \angle R }$

$\implies \bold{180^{\circ} - [ \angle Q + \angle R}}]$

∵

{ ∠P + ∠Q + ∠R = 180°

∠Q + ∠R = 180° - ∠P }

$\implies \bold{\therefore 2\angle QTR = 180^{\circ} - [ 180^{\circ} - \angle P}}$

$\implies \bold{2\angle QTR = 180^{\circ} - 180^{\circ} + \angle P}$

$\implies \bold{2\angle QTR = \angle P}$

$\implies \bold{ \angle QTR = \frac{\angle P }{2}}$

Hence Proved

$\rule{200}2$

Answer 2

Answer:

Given that,

• ∆ PQR
• QR is produced to S.
• Bisector of PQR and PRS meet at T.

To prove,

• ∠QTR = 1/2 ∠QPR

Proof,

We know that, exterior angle is equal to the sum of two opposite interior angles.

So For ∆PQR, ∠PRS = ∠QPR + ∠PQR

Similarly, for ∆TQR, ∠TRS = ∠QRT + ∠TQR

We observe that, ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR.

Using this,

1/2 ∠PRS = ∠QTR + 1/2 ∠PQR

Using ∠PRS = ∠QPR + ∠PQR,

1/2 (∠QPR + ∠PQR) = ∠QTR + 1/2 ∠PQR

Solving it,

1/2 ∠QPR + 1/2 ∠PQR = ∠QTR + 1/2 ∠PQR

1/2 ∠QPR + 1/2 ∠PQR - 1/2 ∠PQR = ∠QTR

1/2 (∠QPR + ∠PQR -∠PQR) = ∠RTQ

1/2 ∠QPR = ∠QTR