Given: ∆ABC, in which D, E and F are the mid points of the sides BC, CA and AB respectively.

To Prove: Each of the triangles AFE, FBD, EDC and DEF are similar to ∆ABC.
Proof: F and E are the mid points of sides AB and AC respectively.
Therefore, FE || BC
[Using mid-point theorem]
Now, in ∆AFE and ∆ABC
∠AFE = ∠B [Corresponding angles]
and    ∠A = ∠A    [Common]
Therefore, by using AA similar condition
∆AFE ~ ∆ABC
Similarly, ∆FBD ~ ∆ABC
and    ∆EDC ~ ∆ABC
∴    DE || AB
⇒    DE || AF    ...(i)
and    DF || AC
⇒    DF || AE    ...(ii)
Comparing (i) and (ii), AEDF is a parallelogram, Similarly BDEF is a parallelogram
Now, in ∆ABC and ∆DEF
∠A = ∠EDF
and    ∠B = ∠DEF
[opposite angles of parallelogram]
Therefore, by using AA similar condition
∆ABC ~ ∆DEF
Thus, each of the triangles AFE, FBD, EDC and DEF arc similar to a ∆ABC.