# Theorem 6.1 - Basic Proportionality Theorem (BPT)

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 :- Δ ABC where DE ∥ BC To Prove :- 𝐴𝐷𝐷𝐵 = 𝐴𝐸𝐸𝐶 Construction :- Join BE and CD Draw DM ⊥ AC and EN ⊥ AB. Proof: Now, Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel BC and DE. ∴ ar (BDE) = ar (DEC) Hence, ar (ADE)ar (BDE) = ar (ADE)ar (DEC) ADDB = AEEC Hence Proved.

Chapter 6 Class 10 Triangles

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.