Question

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

Answer 1

Answer:

PS = PR

Step-by-step explanation:

Missing information RS║ TP

∠QTP = ∠QRS  

∠QPT = ∠QSR      Eq1

PT is the bisector of ∠QPR

=> ∠QPT = ∠TPR      Eq2

∠QTP = ∠TPR + ∠TRP   (  External angle of triangle = sum of opposite internal angle of Triangles)

=> ∠QTP - ∠TRP  = ∠TPR     Eq3

∠PRS = ∠QRS - ∠QRP

=> ∠PRS = ∠QTP - ∠TRP    (  as ∠QTP = ∠QRS  & ∠QRP = ∠TRP  as T lies on QR)

=> ∠PRS =  ∠TPR     from Eq 3

=> ∠PRS = ∠QPT     from eq 2

=> ∠PRS = ∠QSR   from eq 1

∠QSR = ∠PSR  as P lies on QS

=> ∠PRS = ∠PSR

in Δ PRS

∠PRS = ∠PSR

=> PS = PR

QED

Proved

Answer 2

Missing information RS║ TP

∠QTP = ∠QRS  

∠QPT = ∠QSR      Eq1

PT is the bisector of ∠QPR

=> ∠QPT = ∠TPR      Eq2

∠QTP = ∠TPR + ∠TRP   (  External angle of triangle = sum of opposite internal angle of Triangles)

=> ∠QTP - ∠TRP  = ∠TPR     Eq3

∠PRS = ∠QRS - ∠QRP

=> ∠PRS = ∠QTP - ∠TRP    (  as ∠QTP = ∠QRS  & ∠QRP = ∠TRP  as T lies on QR)

=> ∠PRS =  ∠TPR     from Eq 3

=> ∠PRS = ∠QPT     from eq 2

=> ∠PRS = ∠QSR   from eq 1

∠QSR = ∠PSR  as P lies on QS

=> ∠PRS = ∠PSR

in Δ PRS

∠PRS = ∠PSR

=> PS = PR

QED

Proved