To prove two triangles congruent,

We use SSS Criteria – Side Side Side Criteria

In SSS Congruency Criteria,

- All 3 sides of both triangles are equal

__
For example
__

__
__

__
__

Here,

All 3 corresponding sides are equal.

So, the triangles must be congruent

Let’s take some examples

###
**
Are these triangles congruent?
**

**
**

**
**

In ∆ABC and ∆EDF,

AB = ED
*
(
*
*
Both 6 cm
*
*
)
*

BC = DF
*
(
*
*
Both 7 cm
*
*
)
*

AC = EF
*
(Both 8 cm
*
*
)
*

∴ ∆ABC ≅ ∆EDF
*
(
*
*
SSS Congruence Rule)
*

Here,

A⟷E

B⟷D

C⟷F

###
**
Are these triangles congruent?
**

**
**

**
**

In ∆ABC and ∆QPR,

AB = QP
*
(
*
*
Both 14 cm)
*

BC = PR
*
(
*
*
Both 15 cm
*
*
)
*

AC = QR
*
(
*
*
Both 21 cm
*
*
)
*

∴ ∆ABC ≅ ∆QPR
*
(SSS Congruence Rule)
*

Here,

A⟷Q

B⟷P

C⟷R

###
**
Are these triangles congruent?
**

**
**

**
**

In ∆DEF and ∆NML,

DE = NM
*
(
*
*
Both 3.2 cm
*
*
)
*

EF = ML
*
(Both 3 cm
*
*
)
*

DF = NL
*
(
*
*
Both 3.5 cm
*
*
)
*

∴ ∆DEF ≅ ∆NML
*
(
*
*
SSS Congruence Rule)
*

Here,

D⟷N

F⟷L

E⟷M

###
**
Are these triangles congruent?
**

**
**

In ∆ABC and ∆QRP,

AC = QP
*
(
*
*
Both 5 cm
*
*
)
*

BC = RP
*
(
*
*
Both 4cm
*
*
)
*

AB ≠ QR

Since all sides are not equal,

∴ ∆ABC ≇ ∆QRP

###
**
Are these triangles congruent?
**

**
**

In ∆ADB and ∆ADC,

AD = AD
*
(
*
*
Common
*
*
)
*

AB = AC
*
(Both 3.5 cm)
*

DB = DC
*
(Both 2.5 cm)
*

∆ADB ≅ ∆ADC
*
(SSS Congruence Rule)
*

Here,

A⟷A

B⟷C

D⟷D

In Fig, AB =AC and D is the mid-point of (BC) ̅.

(i)State the three pairs of equal parts in ∆ADB and ∆ADC.

(ii)Is ∆ADB ≅ ∆ADC? Give reasons.

(iii)Is ∠B = ∠C? Why?

In ∆ADB and ∆ADC,

AD = AD
*
(
*
*
Common
*
*
)
*

AB = AC
*
(
*
*
Given
*
*
)
*

DB = DC
*
(
*
*
D is the
*
*
mid – point of
*
*
BC
*
*
)
*

∴ ∆ADB ≅ ∆ADC
*
(
*
*
SSS Congruence Rule
*
*
)
*

Also,

∠B = ∠C
*
(By CPCT)
*

Here,

A⟷A

B⟷C

D⟷D

In Fig, AC = BD and AD = BC. Which of the following statements is meaningfully written?

(i)∆ABC ≅ ∆ABD

(ii)∆ABC ≅ ∆BAD

In ∆ABC and ∆BAD,

AB = AB
*
(Common
*
*
)
*

AC = BD
*
(Given
*
*
)
*

AD = BC
*
(Given
*
*
)
*

∴ ∆ABC ≅ ∆BAD
*
(
*
*
SSS Congruence Rule)
*

Here,

A⟷B

C⟷D

B⟷A

Hence (ii) is correct