Theorem 6.2 - Converse of Basic Proportionality Theorem - Chapter 6 Class 10 Triangles

Last updated at April 16, 2024 by Teachoo

Transcript

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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