Now,
∆BDE and ∆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.
Given: Δ ABC and a line DE intersecting AB at D and AC at E,
such that "AD" /"DB" = "AE" /"EC"
To Prove: DE ∥ BC
Construction: Draw DE’ parallel to BC.
Proof:
Since DE’ ∥ BC ,
By Theorem 6.1 :If a line is drawn parallel to one side of a triangle to intersecting other two sides not distinct points, the other two sided are divided in the same ratio.
∴ 𝐴𝐷/𝐷𝐵 = (𝐴𝐸^′)/(𝐸^′ 𝐶)
And given that,
𝐴𝐷/𝐷𝐵 = 𝐴𝐸/𝐸𝐶
From (1) and (2)
(𝐴𝐸^′)/(𝐸^′ 𝐶) = 𝐴𝐸/𝐸𝐶
Adding 1 on both sides
(𝐴𝐸^′)/(𝐸^′ 𝐶) + 1 = 𝐴𝐸/𝐸𝐶 + 1
(𝐴𝐸^′ + 𝐸^′ 𝐶)/(𝐸^′ 𝐶) = (𝐴𝐸 + 𝐸𝐶)/𝐸𝐶
"AE" /"EC" + 1 = "AE′" /"E′C" + 1
("AE" + "EC" )/"EC" = ("AE′" + "E′C" )/"E′C"
"AC" /"EC" = "AC" /"E′C"
EC = E’C
Thus, E and E’ coincides.
Hence, DE ∥ BC.
(𝐴𝐸^′ + 𝐸^′ 𝐶)/(𝐸^′ 𝐶) = (𝐴𝐸 + 𝐸𝐶)/𝐸𝐶
𝐴𝐶/(𝐸^′ 𝐶) = 𝐴𝐶/𝐸𝐶
1/(𝐸^′ 𝐶) = 1/𝐸𝐶
EC = E’C
Thus, E and E’ coincide
Since DE’ ∥ BC
∴ DE ∥ BC.
Hence, proved

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.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Uh-oh

Teachoo answers all your questions if you are a Black user!