Mathematics

In the figure, show that $$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^{\circ}$$.


SOLUTION
We know that the sum of all the angles in triangle $$ACE$$ is $$180^{\circ}$$.
$$\angle A + \angle C + \angle E = 180^{\circ} .. (1)$$
We know that the sum of all the angles in triangle $$BDF$$ is $$180^{\circ}$$.
$$\angle B + \angle D + \angle F = 180^{\circ} .. (2)$$
Now by adding both equations $$(1)$$ and $$(2)$$ we get
$$\angle A + \angle C + \angle E + \angle B + \angle D + \angle F = 180^{\circ} + 180^{\circ}$$
On further calculation
$$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^{\circ}$$
Therefore, it is proved that $$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^{\circ}$$.
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