The midpoint of the side of a triangle are joined together to get four triangles. These four triangles are:

1. not equal to each other

2. congruent to each other

3. not congruent to each other

4. none of these

From figure,

In △ABC,

D, E and F are mid-points of AB, BC and CA respectively.

Now join DE, EF and FD.

To prove :

△ADF ≅ △DBE ≅ △ECF ≅ △DEF

In △ABC,

D and E are midpoints of AB and BC.

∴ DE || AC (By, mid-point theorem) or,

DE || FC ……..(1)

DE || AF ……..(2)

D and F are midpoints of AB and AC.

∴ DF || BC (By, mid-point theorem) or,

DF || EC ……..(3)

DF || BE ……(4)

F and E are midpoints of AC and BC.

∴ FE || AB (By, mid-point theorem) or,

FE || AD ………(5)

FE || DB ………(6)

From (1) and (3) we get,

DE || FC and DF || EC.

Since, opposite sides of a parallelogram are parallel.

∴ DECF is a parallelogram

We know that,

Diagonal of || gm divides it into two congruent triangles.

Diagonal FE divides the parallelogram DECF in two congruent triangles DEF and CEF.

∴ △DEF ≅ △ECF ………(7)

From (2) and (5) we get,

DE || AF and FE || AD.

Since, opposite sides of a parallelogram are parallel.

∴ ADEF is a parallelogram.

We know that,

Diagonal of || gm divides it into two congruent triangles.

Diagonal FD divides the parallelogram in two congruent triangles DEF and AFD.

∴ △DEF ≅ △AFD ………(8)

From (4) and (6) we get,

DF || BE and FE || DB.

∴ DBEF is a parallelogram.

We know that,

Diagonal DE divides the parallelogram in two congruent triangles DEF and DBE.

∴ △DEF ≅ △DBE ………(9)

From equations 7, 8 and 9 we get,

△AFD ≅ △DBE ≅ △ECF ≅ △DEF.

Thus, the four triangles formed are congruent to each other.

Hence, option 2 is the correct option.