Theorem 6.1 Class 10 Maths Explanation with Proof

In this post, we will discuss theorem 6.1 Class 10 Maths which is related to Chapter 6 Triangles Class 10 Mathematics. After understanding theorem 6.1, you can solve the exercise questions given in the NCERT book of Class 10 Maths.

 

 Theorem 6.1 Class 10 Proof

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given 

A triangle ABC in which DE∥BC and DE intersect AB and AC at D and E respectively.

To prove

`\frac{AD}{DE}=frac{AE}{EC}`   

Construction

Join BE and CD. Draw EN⊥AB and DM⊥AC.

Proof 

Since EN⊥AB, therefore EN is the height of triangles ADE and DBE.

Now  , area(∆ADE)= `frac{1}{2} base ×height`

                                 = `frac{1}{2} AD ×EN`

And, area(∆DBE)= `frac{1}{2} base ×height`

                              = `frac{1}{2} BD ×EN`

 ∴  `frac{area(∆ADE)}{area(∆DBE)}= frac{1/2 AD×EN}{1/2 BD×EN}=frac{AD}{BD}`……(i)

Similarly, area (∆ADE)=`frac{1}{2}×AE×DM`

And area(∆DEC)=`frac{1]{2}×EC×DM`

 ∴  `frac{area(∆ADE)}{area(∆DEC)}= frac{1/2 AE×DM}{1/2 EC×DM}=frac{AE}{EC}`……(ii)

But ∆DBE and ∆DEC are on the same base DE and between the same parallels BC and DE,

So,     area(∆DBE)=area( ∆DEC)…………(iii)

Therefore from (i), (ii), and (iii) , we have

`frac{AD}{DB}=frac{AE}{EC]`

Related Topics 

1. Theorem 6.2

2. Theorem 6.3