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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Theorem 6.3 (AAA Criteria) If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangle are similar.Given: Two triangles ∆ABC and ∆DEF such that ∠A = ∠D, ∠B = ∠E & ∠C = ∠F To Prove: ∆ABC ~ ∆DEF Construction: Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ. Proof: In ∆ABC and ∆DPQ AB = DP ∠A = ∠D AC = DQ ⇒ ∆ABC ≅ ∆DPQ ⇒ ∠B = ∠P But, ∠B = ∠E Thus, ∠P = ∠E For lines PQ & EF with transversal PE, ∠ P & ∠ E are corresponding angles, and they are equal Hence, PQ ∥ EF. Since, PQ ∥ EF. By Theorem 6.1, : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 𝐷𝑃/𝑃𝐸 = 𝐷𝑄/𝑄𝐹 𝑃𝐸/𝐷𝑃 = 𝑄𝐹/𝐷𝑄 Adding 1 on both sides. 𝑃𝐸/𝐷𝑃 + 1 = 𝑄𝐹/𝐷𝑄 + 1 (𝑃𝐸 + 𝐷𝑃)/𝐷𝑃 = (𝑄𝐹 + 𝐷𝑄)/𝐷𝑄 𝐷𝐸/𝐷𝑃 = 𝐷𝐹/𝐷𝑄 ⇒ 𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹 And by construction DP = AB and DQ = AC ⇒ 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 Similarly, we can prove that 𝐴𝐵/𝐷𝐸 = 𝐵𝐶/𝐸𝐹 Therefore, 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 = 𝐵𝐶/𝐸𝐹 Since all 3 sides are in proportion ∴ ∆ABC ~ ∆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.