



Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Theorems
Theorem 6.2 - Converse of Basic Proportionality Theorem
Theorem 6.3
AA Similarity Criteria
Theorem 6.4 Important
Theorem 6.5 You are here
Theorem 6.6 Important Deleted for CBSE Board 2024 Exams
Theorem 6.7 Important Deleted for CBSE Board 2024 Exams
Theorem 6.8 Important Deleted for CBSE Board 2024 Exams
Theorem 6.9 Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Theorem 6.5 (SAS Criteria) 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/𝐷𝐹 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. Now, For lines PQ & EF, with transversal PE ∠ P = ∠ E For lines PQ & EF, with transversal QF ∠ Q = ∠ F Now, In ∆ABC and ∆DPQ AB = DP ∠ A = ∠ D AC = DQ ⇒ ∆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