Question

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

Answer 1

Given:
$\angle QPS=\angle RPS$
To Prove:
$\frac{QS}{SR}=\frac{PQ}{PR}$
Construction:
Extend $RP$ to $T$ and
Join $QT$ such that $TQ\parallel PS$
Proof:
Since, $QT\parallel PS$
$\therefore$ $\angle TQP=\angle QPS$ (Alternate Angles)
Also,
$\angle QTP=\angle RPS=\angle QPS$ (Corresponding Angles and $PS$ is the bisector of $\angle QPR$ of $△PQR$)
$\therefore$ $\angle TQP=\angle QTP$
$\therefore TP=QP.......\left(1\right)$
Since, $QT\parallel PS,$ by basic proportionality theorem,
$\therefore$ $\frac{QS}{SR}=\frac{TP}{PR}$
$\therefore$ $\frac{QS}{SR}=\frac{PQ}{PR}$ (From 1)

Hence proved