Question

Prove that the opposite sides of a regular hexagon are parallel.

Answer 1

Construction: in the regular hexagon $ABCDEF$, join $FC$.

In the quadrilateral $ABCF,AF=BC$ and $\angle FAB=\angle CBA={120}^0$

$ABCF$ is an isosceles trapezium

$FC||AB\dots \dots .\left(1\right)$

Similarly $ED||FC\dots \dots .\left(2\right)$

From (1) and (2), we get $AB||ED$

Similarly, $BC||FE$ and $CD||FA$.

Hence, the opposite sides of a regular hexagon are parallel.