In the adjoining figure, show that

In $\Delta A C E$, we have :

$\angle A+\angle C+\angle E=180^{\circ} \ldots(i)$ [Sum of the angles of a triangle]

In $\Delta B D F$, we have:

$\angle B+\angle D+\angle F=180^{\circ} \ldots(i i)$ [Sum of the angles of a triangle]

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

$\angle A+\angle C+\angle E+\angle B+\angle D+\angle F=(180+180)^{\circ}$

$\Rightarrow \angle A+\angle B+\angle C+\angle D+\angle E+\angle F=\mathbf{3 6 0}^{\circ}$