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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Theorem 6.4 (SSS Criteria) : If in two triangles, sides of one triangle are proportional to (i.e., the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangle are similar. Given: Triangle βˆ†ABC and βˆ†DEF such that 𝐴𝐡/𝐷𝐸 = 𝐡𝐢/𝐸𝐹 = 𝐢𝐴/𝐹𝐷 To Prove: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and βˆ†ABC ~ βˆ†DEF Construction: Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ. Proof: Given 𝐴𝐡/𝐷𝐸 = 𝐢𝐴/𝐷𝐹 And DP = AB, DQ = AC 𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹 𝐷𝐸/𝐷𝑃 = 𝐷𝐹/𝐷𝑄 Subtracting 1 on both sides 𝐷𝐸/𝐷𝑃 – 1 = 𝐷𝐹/𝐷𝑄 – 1 (𝐷𝐸 βˆ’ 𝐷𝑃)/𝐷𝑃 = (𝐷𝐹 βˆ’ 𝐷𝑄)/𝐷𝑄 𝑃𝐸/𝐷𝑃 = 𝑄𝐹/𝐷𝑄 𝐷𝑃/𝑃𝐸 = 𝐷𝑄/𝑄𝐹 Using Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side. ∴ PQ βˆ₯ EF. Now, For lines PQ & EF, with transversal PE ∠ P = ∠ E For lines PQ & EF, with transversal QF ∠ Q = ∠ F In βˆ†DPQ and βˆ†DEF ∠P = ∠E ∠Q = ∠F β‡’ βˆ†DPQ ~ βˆ†DEF ∴ 𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹 = 𝑃𝑄/𝐸𝐹 Also, 𝐴𝐡/𝐷𝐸 = 𝐡𝐢/𝐸𝐹 = 𝐢𝐴/𝐹𝐷 & AB = DP & CA = QD Thus, 𝐷𝑃/𝐷𝐸 = 𝐡𝐢/𝐸𝐹 = 𝐷𝑄/𝐹𝐷 From (2) and (3) 𝐡𝐢/𝐸𝐹 = 𝑃𝑄/𝐸𝐹 ∴ BC = PQ In βˆ†ABC and βˆ†DPQ AB = DP AC = DQ BC = PQ β‡’ βˆ†ABC β‰… βˆ†DPQ ∴ ∠B = ∠P ∠C = ∠Q ∠ A = ∠ D But From (1) ∠P = ∠E and ∠Q = ∠F Therefore, ∠B = ∠P = ∠E and ∠C = ∠Q = ∠F Therefore, In Ξ” ABC & Ξ” DEF ∠B = ∠E ∠C = ∠F ∴ βˆ†ABC ~ βˆ†DEF Hence Proved

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Davneet Singh
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.