Question

The sides BC, CA, and AB of an equilateral AABC, are produced in order to form
exterior angles ∆ACD, ∆BAE and CBF. Prove that ∆ACD+∆BAE+∆CBF=360°​

Answer 1

Step-by-step explanation:

To prove: ∠ACD + ∠BAE + ∠CBF = 360°

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

∴ ∠ACD = ∠1 + ∠2 .....(1)

∠BAE = ∠2 + ∠3 .....(2)

and ∠CBF = ∠1 + ∠3 .....(3)

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

∠ACD + ∠BAE + ∠CBF = 2(∠1 + ∠2 +∠3)

= 2 × 180° [ Sum of angles of a triangle is 180°]

= 360°

Hence proved.

Answer 2

Answer:

ABC + ∠CBF = 180° [Linear pair]………..(i) 

∠ACB + ∠ACD = 180° [Linear pair]……(ii)

  ∠BAC + ∠BAE = 180° [Linear pair]…….(iii)

  (i), (ii) and (iii) 

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

180 + 180 + 180  ∠ABC + ∠ACB + ∠BAC + ∠CBF + ∠ACD + ∠BAE = 540°  180° +

∠CBF + ∠ACD + ∠BAE = 540°

  [Sum of the angles of a triangle 180°]  ∠CBF + ∠ACD + ∠BAE = 360°