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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.6, 1 In Fig. 6.56, PS is the bisector of ∠ QPR of Ξ” PQR. Prove that 𝑄𝑆/𝑆𝑅 = 𝑃𝑄/𝑃𝑅 Given : Ξ” PQR and PS is the bisector of ∠QPR i.e. ∠QPS = ∠RPS To Prove: 𝑄𝑆/𝑆𝑅 = 𝑃𝑄/𝑃𝑅 Construction : Draw RT βˆ₯ SP such that RT cuts QP Produced at T. Proof: In Ξ” QRT, RT βˆ₯ SP and PS intersects QT and QR at two distinct points P and Q Therefore, applying Basic Proportionality Theorem in Ξ” QRT QT and QR will be divided in the same ratio 𝑸𝑺/𝑺𝑹 = 𝑷𝑸/𝑷𝑻 Now, we need to prove PT = PR Now RT βˆ₯ SP & PR is the transversal Therefore, Also, Given that PS is the bisector of ∠QPR ∠QPS = ∠RPS ∠1 = ∠2 Putting ∠1 = ∠4 and ∠2 = ∠3 from (2) & (3) ∠4 = ∠3 i.e. ∠PTR = ∠PRT Therefore, PT = PR Putting PT = PR in equation (1) 𝑄𝑆/𝑆𝑅 = 𝑃𝑄/𝑃𝑇 𝑸𝑺/𝑺𝑹 = 𝑷𝑸/𝑷𝑹 Hence Proved.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.