Prove that the line joining the midpoints of the sides of the triangle form four triangles, each of which is similar to the original triangle.
                        

Given: In ΔABC, D, E, F are the midpoints of AB, BC and AC respectively.
 
To prove: ΔABC ~ ΔDEF
ΔABC ~ ΔADF
ΔABC ~ ΔBDE 
ΔABC ~ ΔEFC
Proof: In ΔABC, D and F are mid points of AB and AC respectively. 
∴ DF || BC (midpoint theorem)
In Δ ABC and Δ ADF 
∠A is common; ∠ADF = ∠ABC (corresponding angles)
ΔABC ~ Δ DF (AA similarity) -----(1)
Similarly we can prove ΔABC ~ ΔBDE (AA similarity)-----(2)
ΔABC ~ ΔEFC (AA similarity)-----(3)
In ΔABC and ΔDEF;
since D, E, F are the midpoints of AB, BC and AC respectively,
DF= (1/2) × BC; DE = (1/2) × AC; EF = (1/2) × AB; (midpoint theorem )
∴ AB = BC = CA = 2
 EF    DF    DE
∴ Δ ABC ~ Δ EFD (SSS similarity )------------------(4)
From (1), (2), (3) and (4)
Δ ABC ~ Δ DEF
Δ ABC ~ Δ ADF
Δ ABC ~ Δ BDE 
Δ ABC ~ Δ EFC.