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# In Fig. 6.56, PS is the bisector of angle QPR of Delta PQR. Prove that QS by SR equals PQ by PR

Q1   In Fig. 6.56, PS is the bisector of $\angle QPR \: \: of\: \: \Delta PQR$. Prove that $\frac{QS }{SR } = \frac{PQ }{PR }$
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A line RT is drawn parallel to SP which intersect QP produced at T.

Given: PS is the bisector of $\angle QPR \: \: of\: \: \Delta PQR$.

$\angle QPS=\angle SPR.....................................1$

By construction,

$\angle SPR=\angle PRT.....................................2$   (as PS||TR)

$\angle QPS=\angle QTR.....................................3$   (as PS||TR)

From the above equations, we get

$\angle PRT=\angle QTR$

$\therefore PT=PR$

By construction, PS||TR

In $\triangle$QTR, by Thales theorem,

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

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

Hence proved

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