Theorem 6.3 Class 10 Maths Explanation with Proof

In this post, we will discuss theorem 6.3 Class 10 Maths which is related to Chapter 6 Triangles Class 10 Mathematics. After understanding theorem 6.3, you can solve the exercise questions given in the NCERT book of Class 10 Maths.

 

 Theorem 6.3 Class 10 Proof

(AAA similarity criterion)

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Given 

Two triangles ABC and DEF such that

∠A=∠D,∠B=∠E and ∠C=∠F 

To prove

∆ABC ~ ∆DEF 

Construction 

Cut DP =AB and DQ =AC. Join PQ

Proof 

In ∆ABC and ∆DPQ, we have

AB=DP        (by construction)

AC=DQ      (by construction)

∠A=∠D     (given)

∴∆ABC≅ ∆DPQ (by SAS congruence)

⟹∠B=∠P 

⟹∠E=∠P        [∵∠B=∠E (given)]

But ∠E and ∠P are corresponding angles

Therefore PQ∥EF

⇒`frac{DP}{DE}=frac{DQ}{DF}` [by BPT]

`frac{AB}{DE}=frac{CA}{DF}`  [∵DP=AB and DQ=AC]

∴`frac{AB}{DE}=frac{CA]{FD}`  ………(i)

 ⇒`frac{AB}{DE}=frac{BC}{EF}` ………(ii)

From (i) and (ii), we get

`frac{AB}{DE}=frac{CA}{FD}=frac{BC}{EF}`  

And ∠A=∠D,∠B=∠E and ∠C=∠F (given)

Hence   ∆ABC ~ ∆DEF  proved 

Related Topics 

1. Theorem 6.2

2. Theorem 6.4