[tex]\large\underline{\sf{Given- }}[/tex]

A triangle PQR such that PS is the angle bisector of angle QPR.

[tex] \\ \large\underline{\sf{To\:prove - }}[/tex]

[tex]\sf \: \dfrac{QS}{SR} = \dfrac{PQ}{PR} \\ \\ [/tex]

Construction :-

Through R, draw a line RT parallel to PS intersecting QP at T when produced.

Proof :-

As it is given that, PS is angle bisector of angle QPR.

[tex]\sf \: \angle 1 = \angle 2 - - - (1) \\ \\ [/tex]

By construction, we have PS || RT

[tex]\sf \: \angle 3 = \angle 2 - - - (2) \: \: \{alternate \: interior \}\\ \\ [/tex]

[tex]\sf \: \angle 1 = \angle 4 - - - (3) \: \: \{corresponding \: angles \}\\ \\ [/tex]

So, from equation (1), (2) and (3), we get

[tex]\sf \: \angle 3 = \angle 4\\ \\ [/tex]

[tex]\bf\implies \:PR = PT \\ \\ [/tex]

[ Side opposite to equal angles are equal ]

Now, In triangle QRT

PS || RT

So, By Basic Proportionality Theorem, we have

[tex]\sf \: \dfrac{QS}{SR} = \dfrac{PQ}{PT} \\ \\ [/tex]

[tex]\sf \: \bf\implies \:\dfrac{QS}{SR} = \dfrac{PQ}{PR} \: \: \: \{ \because \: PR = PT \} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Concept Used

Basic Proportionality Theorem,

If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

Answer:

is this faizan? may i ask u that?