The sides AB, BC, CA of a ∠ABC are produced in order, forming exterior angles ∠CBF, ∠ACD and ∠BAE show that ∠ACD + ∠BAE + ∠CBF = 360°

In ΔABC
exterior ∠1 = ∠3 + ∠5 ….(i)
exterior ∠4 = ∠2 + ∠5 ….(ii)
exterior ∠6 = ∠3 +∠2 ….(iii)
Adding equations (i), (ii) and (iii), we get
               ∠1 + ∠4 +∠6 = 2∠3 + 2∠5 + 2∠2
⇒ ∠ACD + ∠BAE + ∠CBF = 2(∠3 + ∠5 +∠2)
      ⇒ But ∠3 + ∠5 +∠2 = 180° ….(Sum of three angles of triangle)
.^.. ∠ACD + ∠BAE + ∠CBF  =  2 x 180° = 360°