Check sibling questions

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

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.