Theorem 7.4 Class 9 Maths Explanation with Proof

In this post, we will discuss theorem 7.4 class 9 Maths which is related to Chapter 7 Triangles Class 9 Mathematics. After understanding  theorem 7.4, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

Theorem 7.4

(SSS congruence rule) : If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Given 

∆ABC and ∆DEF in which

AB=DE, BC=EF

And AC=DF

To prove

∆ABC≅∆DEF 

Construction 

Suppose BC is the longest side of ∆ABC. Draw EG and FG on ∆DEF (corresponding to ∆ABC) such that ∠3=∠1 and ∠4=∠2.Join DG 

Proof 

In ∆ABC and ∆GEF, we have

                                                                              

∠1=∠3                     (by construction)

BC=EF                    (given)

And  ∠2=∠4            ( by construction)

So, ∆ABC ≅∆GEF (ASA congruence rule)

⟹∠A=∠EGF, AB=GE

And, AC=GF

⟹∠A=∠EGF,DE=GE and DF=GF ……(i)

             (∵AB=DE and AC=DF)

Now in ∆EGD, DE=GE [from (1)]

⟹∠5=∠6 

(The angles opposite to equal side of a triangle are equal)

In ∆FDG,  DF=GF [from (1)]

⟹∠7=∠8 

(The angles opposite to equal side of a triangle are equal)

∴∠5+ ∠7=∠6+∠8 

⟹∠EGF=∠EDF 

∠A=∠EDF [from (1)]

⇒∠A=∠D …….(2)

Now in ∆ABC and ∆DEF, we have

AB=DE                    (given)

∠A=∠D                   [from (2)]

AC=DF                   (given)

So, ∆ABC≅∆DEF (SAS congruence rules)

Related Topics

1. Theorem 7.3

2. Theorem 7.5