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# 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.