Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ∠ACD, ∠BAE, and ∠CBF. Show that ∠ACD + ∠BAE + ∠CBF = 360°.

As we know, using theorem (2) i.e. if a side of the triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

So,

∠ ACD = ∠ BAC + ∠ ABC …(1)

∠ BAE = ∠ ABC + ∠ ACB …(2)

∠ CBF = ∠ BAC + ∠ BCA …(3)

Add (1), (2) and (3)

We get,

∠ ACD+∠ BAE+∠ CBF = ∠ BAC+∠ ABC+∠ ABC + ∠ BCA+ ∠ BAC + ∠ BCA

⇒∠ACD+∠BAE+∠CBF = 2(∠ BAC+∠ ABC∠ BCA)

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

∠ABC+ ∠BAC + ∠BCA = 180°

⇒ ∠ACD+∠BAE+∠CBF = 2(180°)

⇒ ∠ACD+∠BAE+∠CBF = 360°

Hence proved.