Question

In the given figure P is any point on side BC of triangle ABC. PQ||BA and PR||CA are drawn. RQ is extended to meet BC produced at S. Prove that SP. SP=SB. BC​

Answer 1

Step-by-step explanation:

1. In ΔBRS

  where PQ║BR , point P and Q divide the line BS and RS in same      ratio.

  We can write this

  $\mathbf{\frac{SP}{SB}=\frac{SQ}{SR}}$          ...1)

2. In ΔRPS

  where QC║RP , point Q and C divide the line PS and RS in same     ratio.

  We can write this

  $\mathbf{\frac{SC}{SP}=\frac{SQ}{SR}}$          ...2)

3. From equation 1) and equation 2), we see that R.H.S is equal in     equation 1) and equation 2). So L.H.S is also equal

   Now we can write

   $\mathbf{\frac{SP}{SB}=\frac{SC}{SP}}$

   SP·SP = SB·SC

   This is it , we have to prove. Please correct your question.

     

Answer 2

Answer:

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