**Question:**

In a convex hexagon, prove that the sum of all interior angle is equal to twice the sum of its exterior angles formed by producing the sides in the same order.

**Solution:**

For a convex hexagon, interior angle $=\left(\frac{2 n-4}{n} \times 90^{\circ}\right)$

For a hexagon, $n=6$

$\therefore$ Interior angle $=\left(\frac{12-4}{6} \times 90^{\circ}\right)$

$=\left(\frac{8}{6} \times 90^{\circ}\right)$

$=120^{\circ}$

So, the sum of all the interior angles $=120^{\circ}+120^{\circ}+120^{\circ}+120^{\circ}+120^{\circ}+120^{\circ}=720^{\circ}$

$\therefore$ Exterior angle $=\left(\frac{360}{n}\right)^{\circ}=\left(\frac{360}{6}\right)^{\circ}=60^{\circ}$

So, sum of all the exterior angles $=60^{\circ}+60^{\circ}+60^{\circ}+60^{\circ}+60^{\circ}+60^{\circ}=360^{\circ}$

Now, sum of all interior angles $=720^{\circ}$

$=2\left(360^{\circ}\right)$

$=$ twice the exterior angles

Hence proved.