The sides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Find the sum of ∠ACD + ∠BAE + ∠CBF.

In △ABC, by using the exterior angle property,

∠ACD = ∠1 + ∠2    ...(i)

∠BAE = ∠2 + ∠3    ...(ii)

∠CBF = ∠1 + ∠3    ...(iii)

Adding these three equations, we get

∠ACD + ∠BAE + ∠CBF = ∠1 + ∠2 + ∠2 + ∠3 + ∠1 + ∠3
⇒ ∠ACD + ∠BAE + ∠CBF = 2(∠1 + ∠2 + ∠3)   ...(iv)

Also in △ABC, by using the angle sum angle property,

∠1 + ∠2 + ∠3 = 180^o

Therefore, from equation (iv),

∠ACD + ∠BAE + ∠CBF = 2 × 180^o

⇒ ∠ACD + ∠BAE + ∠CBF = 360^o