Question

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°.

Answer 1

Given: ∠ACD, ∠BAE and ∠CBF are exterior angles.

To Show: ∠ACD + ∠BAE + ∠CBF = 360°

Proof:

We know that an exterior angle is equal to the sum of its interior opposite angles.

∠1 + ∠2 = ∠ACD … (1)

∠2 + ∠3 = ∠BAE … (2)

∠3 + ∠1 = ∠CBF … (3)

Adding (1), (2) and (3), we get:

∠1 + ∠2 + ∠2 + ∠3 + ∠3 + ∠1 = ∠ACD + ∠BAE + ∠CBF

2(∠1 + ∠2 + ∠3) = ∠ACD + ∠BAE + ∠CBF

2 × 180° = ∠ACD + ∠BAE + ∠CBF (Sum of the angles of a triangle)

∴ ∠ACD + ∠BAE + ∠CBF = 360°