Question:
In the adjoining figure, show that
$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=360^{\circ} .$
Solution:
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:
$\angle A+\angle C+\angle E+\angle B+\angle D+\angle F=(180+180)^{\circ}$
$\Rightarrow \angle A+\angle B+\angle C+\angle D+\angle E+\angle F=\mathbf{3 6 0}^{\circ}$