In the figure, PS is the bisector of \(\triangle\) PQR. Prove that \(QS\over SR\) = \(PQ\over PR\).

Solution :

Given : PQR is a triangle and PS is the bisector of \(\angle\) QPR meeting QR at S.triangle

\(\therefore\)  \(\angle\) QPS = \(\angle\) SPR

To Prove : \(QS\over SR\) = \(PQ\over PR\)

Construction : Draw RT parallel to SP to cut QP produced at T.

Proof : Since PS || TR and PR cuts them, hence we have :

\(\angle\) SPR = \(\angle\) PRT      …….(1)       (alternate angles)triangle

and \(\angle\) QPS = \(\angle\) PTR         ………(2)     (corresponding angles)

but,  \(\angle\) QPS = \(\angle\) PTR        ………(3)      (given)

\(\therefore\)  \(\angle\) PRT = \(\angle\) PTR       [From (1) and (2)]

\(\implies\)  PT = PR           ……….(3)        (sides opposite to equal angles are equal)

Now, in \(\triangle\) QRT, we have :

SP || RT               (by construction)

\(\therefore\)  \(QS\over SR\) = \(PQ\over PT\)       (by basic proportionality)

or          \(QS\over SR\) = \(PQ\over PR\)           (from 3)

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