Theorem 6.4 Class 9 Maths Explanation with Proof

Theorem 6.4 Class 9 Maths Explanation with Proof

In this post, we will discuss theorem 6.4 class 9 Maths which is related to Chapter 6 Lines and Angles Class 9 Mathematics. After understanding theorem 6.4, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

Theorem 6.4 Class 9 Maths

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Given 

Lines AB and CD are parallel and a transversal r cuts Ab and CD at points P and Q respectively. Thus two pairs of interior angles(∠4,∠5) and (∠3,∠6) are formed.

 To prove :

Each pair of interior angles on the same side of the transversal is supplementary, ie. ∠3+∠6=180°.

 

Poof :

Here ray PQ stands on line AB

So, ∠4+∠3=180° (By linear pair axiom)……(i)

But ∠3=∠5   (alternate interior angles)

On putting ∠3=∠5   in equation (i) , we get

∠4+∠5=180°   

Again, ray PQ stands on line CD

Then, ∠5+∠6=180℃   (By linear pair axiom)……(ii)

But ∠5=∠3   (alternate interior angles)

On putting ∠5=∠3   in equation (ii), we get

∠3+∠6=180°   

Hence ∠3+∠6=180°   and ∠4+∠5=180° 

Hence proved 

Related Topics for You

1.    Next theorem – Theorem 6.5 Class 9 Maths

2.    Previous theorem –Theorem 6.3 Class 9 Maths

3.     NCERT Solution of Exercise  6.2