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

In ΔABC, D, E and F are the mid-points of sides AB, BC and AC respectively.
To Prove: ΔADF ≅ ΔDBE ≅ ΔECF ≅ ΔFDE
Proof: In ΔABC, F is the mid-point of AC and D is 
the mid-point of AB.
.^.. FD = CB and FD ïï CB (mid-point theorem) ⇒ FD = CE and FD ïï CE...(i)
Similarly,
DE = FC and DE ïï FC..(ii)
and FE = DB and FE ïï DB...(iii)
.^.. From (i), (ii) and (iii), we get
ADEF, DBEF and DECF are parallelograms. 
The diagonal of a parallelogram divide it into two triangles which will be  congruent to each other.
.^.. ΔDEF ≅ ΔADF..(iv)
.^.. ΔDEF ≅ ΔDBE...(v)
.^.. ΔDEF ≅ ΔFEC  ...(vi)
From (iv), (v) and (vi) we get,  
ΔADF ≅ ΔDBE ≅ ΔECF ≅ ΔDEF