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Theorem 6.5 (SAS Similarity) - Class 10 - If one angle of triangle is equal to one angle of the other.jpg

2 Theorem 6.5 Subtracting 1 on both sides DEDP - 1 DF DQ - 1.jpg
3 Theorem 6.5 if line divides any two sides of triangle.jpg4 Theorem 6.5 ABC DEF (by AA simillarity criteria hence proved.jpg5 Theorem 6.5 ABC DEF (by AA simillarity criteria hence proved.jpg

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Theorem 6.5 (SAS Similarity) If one angle of a triangle is equal to one angle of the other triangle and sides including these angles are proportional then the triangles are similar. Given :- Two triangles ∆ABC and ∆DEF such that ∠A = ∠D AB﷮DE ﷯ = AC﷮DF﷯ To Prove : ∆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 DF − 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 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 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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