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Theorem 6.4 - Class 10th - If sides are in same ratio, two triangles are similar. - Theorems

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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 :- Two triangle ∆ABC and ∆DEF such that AB﷮DE ﷯ = BC﷮EF﷯ = CA﷮FD﷯ To Prove :- ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and ∆ABC ~ ∆DEF Construction :- Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ. Proof :- Given AB﷮DE ﷯ = CA﷮FD﷯ 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 − DP﷮DP﷯ = DF − 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 ∥ EF. Thus, ∠P = ∠E and ∠Q = ∠F Now, In ∆DPQ and ∆DEF ∠P = ∠E ∠Q = ∠F ⇒ ∆DPQ ~ ∆DEF ∴ DP﷮DE ﷯ = DQ﷮DF﷯ = PQ﷮PF﷯ Also, AB﷮DE ﷯ = BC﷮EF﷯ = CA﷮FD﷯ and AB = DP & CA = QD ⇒ DP﷮DE ﷯ = BC﷮EF﷯ = QD﷮FD﷯ From (2) and (3) BC﷮EF﷯ = PQ﷮EF﷯ ⇒ BC = PQ In ∆ABC and ∆DPQ AB = DP AC = DQ BC = PQ ⇒ ∆ABC ≅ ∆DPQ ∴ ∠B = ∠P ∠C = ∠Q But From (1) ∠P = ∠E and ∠Q = ∠F Therefore, ∠B = ∠P = ∠E and ∠C = ∠Q = ∠F Therefore, In Δ ABC & Δ DEF ∠B = ∠E ∠C = ∠F ∴ ∆ABC ~ ∆DEF Hence Proved

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Davneet Singh
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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