

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Theorems
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 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
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