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Last updated at Aug. 13, 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: 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.

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

Theorem 6.5

Theorem 6.6 Not in Syllabus - CBSE Exams 2021

Theorem 6.7 Important Not in Syllabus - CBSE Exams 2021

Theorem 6.8

Theorem 6.9

Chapter 6 Class 10 Triangles

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.