Theorems

Theorem 6.1 - Basic Proportionality Theorem (BPT)
Important

Theorem 6.2 - Converse of Basic Proportionality Theorem You are here

Theorem 6.3

AA Similarity Criteria

Theorem 6.4 Important

Theorem 6.5

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

Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel lines BC and DE. ∴ ar (BDE) = ar (DEC) Hence, "ar (ADE)" /"ar (BDE)" = "ar (ADE)" /"ar (DEC)" "AD" /"DB" = "AE" /"EC" Hence Proved. Given: Δ ABC and a line DE intersecting AB at D and AC at E, such that "AD" /"DB" = "AE" /"EC" To Prove: DE ∥ BC Construction: Draw DE’ parallel to BC. Proof: Since DE’ ∥ BC , By Theorem 6.1 :If a line is drawn parallel to one side of a triangle to intersecting other two sides not distinct points, the other two sided are divided in the same ratio. ∴ 𝐴𝐷/𝐷𝐵 = (𝐴𝐸^′)/(𝐸^′ 𝐶) And given that, 𝐴𝐷/𝐷𝐵 = 𝐴𝐸/𝐸𝐶 From (1) and (2) (𝐴𝐸^′)/(𝐸^′ 𝐶) = 𝐴𝐸/𝐸𝐶 Adding 1 on both sides (𝐴𝐸^′)/(𝐸^′ 𝐶) + 1 = 𝐴𝐸/𝐸𝐶 + 1 (𝐴𝐸^′ + 𝐸^′ 𝐶)/(𝐸^′ 𝐶) = (𝐴𝐸 + 𝐸𝐶)/𝐸𝐶 "AE" /"EC" + 1 = "AE′" /"E′C" + 1 ("AE" + "EC" )/"EC" = ("AE′" + "E′C" )/"E′C" "AC" /"EC" = "AC" /"E′C" EC = E’C Thus, E and E’ coincides. Hence, DE ∥ BC. (𝐴𝐸^′ + 𝐸^′ 𝐶)/(𝐸^′ 𝐶) = (𝐴𝐸 + 𝐸𝐶)/𝐸𝐶 𝐴𝐶/(𝐸^′ 𝐶) = 𝐴𝐶/𝐸𝐶 1/(𝐸^′ 𝐶) = 1/𝐸𝐶 EC = E’C Thus, E and E’ coincide Since DE’ ∥ BC ∴ DE ∥ BC. Hence, proved