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.jpg 4 Theorem 6.5 ABC DEF (by AA simillarity criteria hence proved.jpg 5 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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