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Criteria for Similarity of Triangles

 

Characteristic Property 1

Theorem (AAA Similarity)  
If in two triangles corresponding angles are equal, i.e. if the two triangles are equiangular, then the triangles are similar.

Given 
Two ΔABC and ΔDEF in which ∠A = ∠D, ∠B = ∠E and ∠C = ∠F (see Fig.)

To Prove
ΔABC ~ ΔDEF

Construction 
Mark point M on DE and N on DF such that AB = DM and AC = DN. Join M to N.

Proof  
There are three cases:

Case (i) 

 

AB=DE (M lies on E)
∴ ΔABC ≅ ΔDEF  [ASA]
∴ AB=DE

BC=EF and AC=DF [C.P.C.T.E.]

Therefore, N coincides with F,


and ∠A=∠D, ∠B=∠E
and ∠C=∠F (given)

Hence, ΔABC ~ ΔDEF.

Case (ii)

AB
In DABC and DDMN, 
AB=DM, 
AC=DN [by construction] 
and ∠A=∠D (Given)


∴ DABC ≅ DDMN [By SAS rule]
∴ ∠B = ∠DMN (C.P.C.T.E.)

But, ∠B = ∠DEF   
∴ ∠DMN = ∠DEF

Hence, MN || EF [Corresponding angles of two lines are equal]
∴ [B.P. Theorem]

⇒     ------ (i) [AB = DM, AC = DN]

Similarly   ------(ii),

From the equation (i) and (ii),
We get  

Also, ∠A = ∠D, ∠B = ∠E
And ∠C = ∠F (given)
ΔABC ~ ΔDEF.

Case (iii)

 


AB > DE. Here M lies on DE (produced).
In ΔABC and ΔDMN,
AB = DM, AC = DN and ∠A = ∠D   
             

∴ DABC ≅ DDMN
∴ ∠B = ∠M

But ∠B = ∠DEF
   
∴ ∠DMN = ∠DEF
∴ EF || MN


But DM = AB and DN = AC

∴  ----(i)

Similarly we can prove,  

 -------(ii),

From equation (i) and (ii), we get



also ∠A=∠D, ∠B=∠E and ∠C=∠F (given)

∴ ΔABC ~ ΔDEF
 



 

Corollar

 

The sum of the three angles in a triangle is 180°, therefore, if two angles of a triangle are equal to two angles of another triangle then they will be similar.

Characteristic Property 2
 Theorem SSS Similarity
If corresponding sides of two triangles are proportional then, the triangles are similar.

Given   
Two triangles ABC and DEF, such that


To Prove  ΔABC ~ ΔDEF

Construction

Mark point M on DE and N on DF such that DM=AB and DN=AC, Join M to N.

Proof
Since  and  AB=DM, AC=DN
∴  ⇒ MN || EF  

Hence, ∠DMN = ∠DEF and ∠DNM = ∠DFE  


∴ DDMN ~ DDEF  

⇒   
or

Hence, MN =BC  
Again in ΔABC and ΔDMN, AB = DM, AC = DN and BC = MN
∴ ΔABC ≅ ΔDMN [SSS rule]
since  ΔDMN ~ ΔDEF  
∴ ΔABC ~ ΔDEF

Characteristic Property 3

Theorem SAS Similarity
Two triangles will be similar, if one pair of corresponding sides is proportional and the included angles are equal.

Given 
 


ΔABC and ΔDEF such that,
 and ∠A = ∠D

To Prove ΔABC ~ ΔDEF

Construction  
Mark a point M on DE and N on DF such that DM=AB and DN = AC. Join MN.

Proof  
In the ΔABC and ΔDMN, AB = DM, AC = DN
And ∠A = ∠D
ΔABC ≅ ΔDMN [SAS rule]
ΔABC ~ ΔDMN [Congruent triangles are similar] (i)

we have,  
  (given) 
⇒ 
⇒ MN || EF [Converse of basic proportionality theorem]
...      ∠DMN = ∠DEF
and  ∠DNM = ∠DFE
... ΔDMN ~ ΔDEF (ii)
From equations (i) and (ii) we get,
ΔABC ~ ΔDEF.
 





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