If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
Let, ABC be a triangle and AB, BC and AC produced to D, E and F respectively.
∠A + ∠B + ∠C = 180o (i)
∠CBD = ∠C + ∠A (Exterior angle theorem) (ii)
∠ACE = ∠A + ∠B (Exterior angle theorem) (iii)
∠BAF = ∠B + ∠C (Exterior angle theorem) (iv)
Adding (ii), (iii) and (iv) we get
∠CBD + ∠ACE + ∠BAF = 2∠A + 2∠B + 2∠C
∠CBD + ∠ACE + ∠BAF = 2 (∠A + ∠B + ∠C)
∠CBD + ∠ACE + ∠BAF = 2 * 180o
∠CBD + ∠ACE + ∠BAF = 360o
Thus, sum of all three exterior angles is 360o.