Theorems

Theorem 6.1 - Basic Proportionality Theorem (BPT)
Important

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 2025 Exams

Theorem 6.7 Important Deleted for CBSE Board 2025 Exams

Theorem 6.8 Important Deleted for CBSE Board 2025 Exams

Theorem 6.9 Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 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