In figure 3.62, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR

Given: PT is angle bisector of ∠QPR

⇒ ∠QPT = ∠RPT

A line through R parallel to PT intersects ray QP at S

RS || PT

To Prove: PS = PR

Proof:

PT is angle bisector of ∠QPR

⇒ ∠QPT = ∠RPT

∠QPR = ∠QPT + ∠RPT

∠QPR = 2∠RPT (1)

RS || PT, PR is the transversal

So, ∠RPT = ∠PRS [Alternate interior angles] (2)

For ΔPRS ∠RPQ is the remote exterior angle.

∠PSR + ∠PRS = ∠QPR

Substituting (1) and (2) in the above equation

∠RPT + ∠PSR = 2∠RPT

⇒ ∠PSR = ∠RPT (3)

From (2) and (3)

∠PRS = ∠PSR

⇒ PS = PR [Sides opposite to equal angles are equal]