Theorem 6.2 Class 10 Maths Explanation with Proof

In this post, we will discuss theorem 6.2 Class 10 Maths which is related to Chapter 6 Triangles Class 10 Mathematics. After understanding theorem 6.2, you can solve the exercise questions given in the NCERT book of Class 10 Maths.

 

 Theorem 6.2 Class 10 Proof

(Converse of Thales Theorems)

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given 

A ∆ABC and a line l intersecting AB at D and AC at E, such

`\frac{AD}{DE}=frac{AE}{EC}`

To prove

DE∥BC 

Proof 

 If possible, let DE is not parallel to BC, then there must be another line through D, which is parallel to BC.

 Therefore, by the Basic proportionality theorem, we have,

`frac{AD}{DB}=frac{AF}{FC}`     ………(1)

But 

`frac{AD}{DB}=frac{EC}{EC}`  (given)   ………….(2)

From (1) and (2), we get 

`frac{AF}{FC}=frac{AE}{EC}`

    

 ⟹  `frac{AF}{FC}+1=frac{AE}{EC}+1` [Adding 1 on both sides]

⟹ `frac{AF+FC}{FC}=frac{AE+EC}{EC}` 

⟹ `frac{AC}{FC}=frac{AC]{EC}`   

⟹  `frac{1}{FC}=frac{1}{EC}` 

⟹FC=EC 

This is possible only when E and F coincide.

Hence, DE∥BC  proved.

Related Topics 

1. Theorem 6.2

2. Theorem 6.1