Theorem 6.3 - (AAA Similarity) If in two traingles, corresponding angles are equal, then their corresponding.jpg

2 Theorem 6.3 - But they are corresponding angles for lines AB & CD with transversal PS .jpg
3 Theorem 6.3 - Since PQ EF by theorem 6.1 if line is drawn parallel to one side of triangle .jpg 4 Theorem 6.3 - Similarly we can Prove that ABDE = .jpg

  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 AC = DQ ∠A = ∠D ⇒ ∆ABC ≅ ∆DPQ ⇒ ∠B = ∠P But ∠B = ∠E Thus, ∠P = ∠E But they are corresponding angles. For lines AB & CD with transversal PS, corresponding angles 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﷯ & ∆ABC ~ ∆DEF Hence Proved.

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