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Theorem 6.4 - Class 10 - If sides are in same ratio (SSS).jpg

2 Theorem 6.4 - Subtracting 1 on both sides DEDP - 1 = DFPQ - 1.jpg
3 Theorem 6.4 - Using theorem 6.2 if line divides any two sides of  triangle in the same ratio.jpg4 Theorem 6.4 - from 2 and 3 BCEF = PQEF BC = PQ.jpg5 Theorem 6.4 - Therefore B = P = E  and C = Q = F.jpg6 Theorem 6.4 - ABC - DEF by AA Similarity Criteria Hence proved.jpg

  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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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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