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

2 Theorem 6.1 - Now ar ADE = 12 base  height = 12 AD EN.jpg
3 Theorem 6.1 - BDE and DEC are on the same base DE and between the same parllel BC and DE.jpg

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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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 :- 𝐴𝐷﷮𝐷𝐵﷯ = 𝐴𝐸﷮𝐸𝐶﷯ 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)﷯ AD﷮DB﷯ = AE﷮EC﷯ Hence Proved.

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