Theorem 6.7 Class 9 Maths Explanation with Proof

Theorem 6.7 Class 9 Maths Explanation with Proof

In this post, we will discuss theorem 6.7 class 9 Maths which is related to Chapter 6 Lines and Angles Class 9 Mathematics. After understanding theorem 6.7, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

Theorem 6.7 Class 9 Maths

The sum of the angles of a triangle is 180°

Given 

A ∆ABC  whose angles are represented by 1,2 and 3,i.e ∠B=∠1,∠C=∠2, and ∠A=∠3.

To prove

Sum of the angles= 180° i.e., ∠1+∠2+∠3=180°

construction: Through A draw a line PQ parallel to BC.

Proof 

Here PQ||BC and AB is the transversal.

∠1=∠4(Alternate interior angles) ……. (i)

Again, PQ||BC and AC is the transversal 

∠2=∠5(Alternate interior angles) ……. (ii)

On adding equations (i) and (ii), we get

∠1+∠2=∠4+∠5

On adding ∠3 in both sides, we get

∠1+∠2+∠3=∠3+∠4+∠5

∠1+∠2+∠3=180°

Since sum of all the angles formed on the same side of a line at a given point on the line is 180°

So, the sum of three angles of a triangle is 180°

Hence proved

Related Topics for You

1.    Next theorem – Theorem 6.8 Class 9 Maths

2.    Previous theorem – Theorem 6.6 Class 9 Maths

3.     NCERT Solution of Exercise  6.2