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Theorem 6.1 - Basic Proportionality if line is drawn parallel to one side of triangle to intersect the other two side .jpg

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. Class 10
  2. Important Questions for Exam - Class 10
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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 :- 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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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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