In Figure, PS is the bisector of \(\angle\) QPR of \(\triangle\) PQR. Prove that \(\frac{QS}{PQ} = \frac{SR}{PR} \)

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given, PS is the angle bisector of \(\angle\) QPR. Therefore,

\(\angle\) QPS = \(\angle\) SPR ………..(i)
As per the constructed figure,
\(\angle\) SPR=\(\angle\) PRT(Since, PS||TR)……………(ii)
\(\angle\) QPS = \(\angle\) QRT(Since, PS||TR) …………..(iii)

From the above equations, we get,
\(\angle\) PRT=\(\angle\) QTR
Therefore,
PT=PR

In \(\triangle\) QTR, by basic proportionality theorem,
\(\frac{QS}{SR} = \frac{QP}{PT} \)
Since, PT=TR

Therefore,
\(\frac{QS}{SR} = \frac{PQ}{PR} \)
Hence, proved.