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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 \triangleQTR, by Thales theorem,

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

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

Hence proved 

 

 

 

 

 

 

 

 

 

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