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Theorems

Theorem 6.1 - Basic Proportionality Theorem (BPT)
Important
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Theorem 6.2 - Converse of Basic Proportionality Theorem

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

Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio. Given: Ξ ABC where DE β₯ BC To Prove: π΄π·/π·π΅ = π΄πΈ/πΈπΆ Construction: Join BE and CD Draw DM β₯ AC and EN β₯ AB. Proof: Now, Now, ar (ADE) = 1/2 Γ Base Γ Height = 1/2 Γ AE Γ DM ar (DEC) = 1/2 Γ Base Γ Height = 1/2 Γ EC Γ DM Divide (3) and (4) "ar (ADE)" /"ar (DEC)" = (1/2 " Γ AE Γ DM" )/(1/2 " Γ EC Γ DM " ) "ar (ADE)" /"ar (DEC)" = "AE" /"EC" Now, βBDE and βDEC are on the same base DE and between the same parallel lines BC and DE. β΄ ar (BDE) = ar (DEC) Hence, "ar (ADE)" /"ar (BDE)" = "ar (ADE)" /"ar (DEC)" "AD" /"DB" = "AE" /"EC" Hence Proved.