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Theorem 6.3 (AAA Criteria) - Class 10th - If corresponding angles are equal, triangles are similar - Theorems

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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 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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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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