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

$PS$ is the bisector of $\angle QPR$ of $∆PQR$.

To do:

We have to prove that $\frac{QS}{SR}=\frac{PQ}{PR}$

Draw $RT \| PS$,meeting $QP$ produced at $T$.

$PS \| RT$

This implies,

$\angle 2=\angle 3$             (Alternate angles)

$\angle 1=\angle 4$        (Corresponding angles)

$\angle 1 = \angle 2$       (AD is the bisector of $\angle A$)

Therefore,

$\angle 3 =\angle 4$

This implies,

 $PT=PR$

In $\triangle QRT$,

$\frac{QS}{SR}=\frac{PQ}{PT}$

$\frac{QS}{SR}=\frac{PQ}{PR}$

Hence proved.

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Updated on: 10-Oct-2022

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