Question

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

Answer 1

We know that the sum of all the angles in triangle $ACE$ is ${180}^{\circ }$.
$\angle A+\angle C+\angle E={180}^{\circ }\dots ..\left(1\right)$
We know that the sum of all the angles in triangle $BDF$ is ${180}^{\circ }$.
$\angle B+\angle D+\angle F={180}^{\circ }\dots ..\left(2\right)$
Now by adding both equations $\left(1\right)$ and $\left(2\right)$ 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 $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F={360}^{\circ }$.