The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively.

(d) 360°

We have :

$\angle 1+\angle B A E=180^{\circ} \ldots($ i $)$

$\angle 2+\angle C B F=180^{\circ} \quad \ldots($ ii $)$ and

$\angle 3+\angle A C D=180^{\circ} \quad \ldots($ iii $)$

Adding $(i),(i i)$ and $(i i i)$, we get:

$(\angle 1+\angle 2+\angle 3)+(\angle B A E+\angle C B F+\angle A C D)=540^{\circ}$

$\Rightarrow 180^{\circ}+\angle B A E+\angle C B F+\angle A C D=540^{\circ} \quad\left[\because \angle 1+\angle 2+\angle 3=180^{\circ}\right]$

$\Rightarrow \angle B A E+\angle C B F+\angle A C D=360^{\circ}$