Question

PS is the bisector of angle QPR of a triangle pqr prove that QS℅SR= PQ℅SR

Answer 1

Draw RX parallel to PS to meet QP produced at X
Let angle QPS=angle 1,angleSPR=angle2, anglePXR =angle,anglePRX=angle3.

Angle2=angle3. (alternate angle)
Angle=angle4. (corresponding angle)

But,angle1=angle2. (since PS is angle bisector)
Therefore, angle3=angle4 => PX=PR

Now, in ∆QRX,we have RX parallel to PS

=> QS/QP =SR/PX

=>QS/SR = QP/PX => QS/SR =PQ/PR. (since,PR=PX proved above)

Hence,QS/SR =PQ/PR

Answer 2

Given: A triangle PQR in which PS is the internal bisector of angle P.

R.T.P: QS/SR=PQ/PR

Construction: Draw RE parallel PS to meet QP produced in E.

Proof:
Here we have,
ER || PS
Then,
angle 2 = angle 3.........1 (alternative angles)
and,
angle 1 = angle 4....... 2 (corresponding angles)
Given that PS in an angular bisector
then,we have
angle 1 = angle 2........3
By eq1 , eq2 , eq3 we get,
angle 3 = angle 4
Then,
PR=PE...........4 (sides opposite to equal angles)
In triangle QRE we have,
RE || PS
By basic proportionality theorem we get,
QS/SR = QP/PE

QS/SR = QP/PR

QS/SR = PQ/PR

Hence proved.