Question

In a regular hexagon $ABCDEF$ with center at $O$ (origin). Prove
that AB+AC+AD+AE+AF=3AD.

Answer 1

Dividing hexagon into $4$ triangles by joining $AC,AD$ and $AE.$ Considering triangle $ACD$ we get $AD=AC+CD⟶\left(1\right)$

from $△ADE$ $AD=AE+ED⟶\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 2AD=AC+CD+AE+ED$

Now, $AF=CD$ because its regular hexagon therefore it has equal and parallel opposite sides.

Similarly $ED=AB$

substituting $ED=AB$ in equation

$\Rightarrow 2AD=AB+AC+AE+AF$

Adding one more $AD$ both sides and gives desired result.

Hence, proved.