Prove that four triangles formed by joining in pairs, the mid-points of the sides of a triangle are congruent to each other.

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 or,

DE || FC …….(i)

or DE || AF ………(ii)

D and F are midpoints of AB and AC

∴ DF || BC or,

DF || EC …….(iii)

or DF || BE ……..(iv)

F and E are midpoints of AC and BC

∴ FE || AB or,

FE || AD …….(v)

or FE || DB (vi)

From (i) and (iii) we get,

DE || FC and DF || EC.

∴ DECF is a parallelogram.

We know that,

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

∴ △DEF ≅ △ECF …….(1)

From (ii) and (v) we get,

DE || AF and FE || AD.

∴ ADEF is a parallelogram.

We know that,

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

∴ △DEF ≅ △AFD …….(2)

From (iv) and (vi) 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 …….(3)

Using equations 1, 2 and 3 we get,

△ADF ≅ △DBE ≅ △ECF ≅ △DEF.

Hence, proved that four triangles formed by joining in pairs, the mid-points of the sides of a triangle are congruent to each other.