The sides BC, CA and AB of ΔABC have been produced to D, E and F respectively as shown in the figure, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Then, ∠ACD + ∠BAE + ∠CBF = ?

Question

The sides BC, CA and AB of ΔABC have been produced to D, E and F respectively as shown in the figure, forming exterior angles ∠ ACD, ∠ BAE and ∠ CBF. Then, ∠ ACD + ∠ BAE + ∠ CBF = ?

Philoid · Accepted Answer

In Δ ABC,

we have CBF = 1 + 3 ...(i) [exterior angle is equal to the sum of opposite interior angles] Similarly, ACD = 1 + 2 ...(ii)

and BAE = 2 + 3 ...(iii)

On adding Eqs. (i), (ii) and (iii),

we get CBF + ACD + BAE =2[ 1 + 2 + 3] = 2 × 180° = 4 × 90°

[by angle sum property of a triangle is 180°] CBF + ACD + BAE = 4 right angles

Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so formed is equal to four right angles = 360°