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.

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.