# 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 Î”ABC and Î”DEF in which âˆ A = âˆ D, âˆ B = âˆ E and âˆ C = âˆ F (see Fig.)

**To Prove**

Î”ABC ~ Î”DEF

**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)

âˆ´ Î”ABC â‰… Î”DEF [ASA]

âˆ´ AB=DE

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

Therefore, N coincides with F,

and âˆ A=âˆ D, âˆ B=âˆ E

and âˆ C=âˆ F (given)

Hence, Î”ABC ~ Î”DEF.

**Case (ii)**

AB

In DABC and DDMN,

AB=DM,

AC=DN [by construction]

âˆ´ DABC â‰… DDMN [By SAS rule]

âˆ´ âˆ B = âˆ DMN (C.P.C.T.E.)

But, âˆ B = âˆ DEF

âˆ´ âˆ DMN = âˆ DEF

Hence, MN || EF [Corresponding angles of two lines are equal]

âˆ´ [B.P. Theorem]

â‡’ ------ (i) [AB = DM, AC = DN]

Similarly ------(ii),

From the equation (i) and (ii),

We get

Also, âˆ A = âˆ D, âˆ B = âˆ E

And âˆ C = âˆ F (given)

Î”ABC ~ Î”DEF.

**Case (iii)**

AB > DE. Here M lies on DE (produced).

In Î”ABC and Î”DMN,

AB = DM, AC = DN and âˆ A = âˆ D

âˆ´ DABC â‰… DDMN

âˆ´ âˆ B = âˆ M

But âˆ B = âˆ DEF

âˆ´ âˆ DMN = âˆ DEF

âˆ´ EF || MN

âˆ´

But DM = AB and DN = AC

âˆ´ ----(i)

Similarly we can prove,

-------(ii),

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

also âˆ A=âˆ D, âˆ B=âˆ E and âˆ C=âˆ F (given)

âˆ´ Î”ABC ~ Î”DEF

# Corollar

The sum of the three angles in a triangle is 180Â°, 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**Î”ABC ~ Î”DEF

**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

âˆ´ â‡’ MN || EF

Hence, âˆ DMN = âˆ DEF and âˆ DNM = âˆ DFE

âˆ´ DDMN ~ DDEF

â‡’

or

Hence, MN =BC

Again in Î”ABC and Î”DMN, AB = DM, AC = DN and BC = MN

âˆ´ Î”ABC â‰… Î”DMN [SSS rule]

since Î”DMN ~ Î”DEF

âˆ´ Î”ABC ~ Î”DEF

**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**

Î”ABC and Î”DEF such that,

and âˆ A = âˆ D

**To Prove **Î”ABC ~ Î”DEF

**Construction**

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

**Proof**

In the Î”ABC and Î”DMN, AB = DM, AC = DN

And âˆ A = âˆ D

Î”ABC â‰… Î”DMN [SAS rule]

Î”ABC ~ Î”DMN [Congruent triangles are similar] (i)

we have, (given)

â‡’

â‡’ MN || EF [Converse of basic proportionality theorem]

... âˆ DMN = âˆ DEF

and âˆ DNM = âˆ DFE

... Î”DMN ~ Î”DEF (ii)

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

Î”ABC ~ Î”DEF.