In the figure,$PS$ is the bisector of $\angle QPR$ of $\Delta PQR$. Prove that $\frac{QS}{SR} = \frac{PQ}{PR}$.

(N/A) Construction: Draw a line segment $RT$ parallel to $PS$ which intersects the extended line segment $QP$ at point $T$.
Given that,$PS$ is the angle bisector of $\angle QPR$.
Therefore,$\angle QPS = \angle SPR \quad \dots(1)$
By construction,$PS \parallel TR$ and $QT$ is a transversal.
Therefore,$\angle QPS = \angle QTR$ (Corresponding angles) $\quad \dots(2)$
Also,$PS \parallel TR$ and $PR$ is a transversal.
Therefore,$\angle SPR = \angle PRT$ (Alternate interior angles) $\quad \dots(3)$
From equations $(1)$,$(2)$,and $(3)$,we get:
$\angle QTR = \angle PRT$
In $\Delta PTR$,since $\angle QTR = \angle PRT$,the sides opposite to these angles are equal.
Therefore,$PT = PR \quad \dots(4)$
In $\Delta QTR$,since $PS \parallel TR$,by the Basic Proportionality Theorem $(BPT)$:
$\frac{QS}{SR} = \frac{QP}{PT}$
Substituting $PT = PR$ from equation $(4)$ into the above expression:
$\frac{QS}{SR} = \frac{PQ}{PR}$
Hence,it is proved.