Theorem 6.1 - Basic Proportionality Theorem (BPT) - Chapter 6 Class 10 Triangles

Last updated at July 14, 2026 by

Theorem 6.1 - Basic Proportionality Theorem (BPT) - Chapter 6 Class 10

Theorem 6.1 - Basic Proportionality Theorem (BPT) - Chapter 6 Class 10 Triangles - Part 2
Theorem 6.1 - Basic Proportionality Theorem (BPT) - Chapter 6 Class 10 Triangles - Part 3

 

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Transcript

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 βˆ₯ BC To Prove: 𝐴𝐷/𝐷𝐡 = 𝐴𝐸/𝐸𝐢 Construction: Join BE and CD Draw DM βŠ₯ AC and EN βŠ₯ 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, βˆ†BDE and βˆ†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.

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