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I need a theoretical proof that each angle of a regular hexagon is $120^\circ$.

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Draw the five radiuses from the hexagonal's center to its vertices. You get 6 congruent isosceles triangles whose basis angle's equals $\,x=\,$ half our wanted angle..

But then the central angle in each triangle equals $\,180^\circ-2x\,$ , so

$$6(180^\circ-2x)=360^\circ\Longrightarrow 180^\circ-2x=60^\circ\Longrightarrow 2x=120^\circ$$

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Sum of the internal angles of a regular Hexagon is $(2.6-4)*90=720 $ As there are 6 angles so each angle must be $720/6=120$

Another Perspective:In a regular Hexagon there are 6 Triangles each having the sum of the angles as $180$ degrees.So the sum of the Angles of all the 6 triangles is $180*6$ but this contains the central angle of $360$ degree so the sum of the internal angles of the hexagon is $180.6-360=720$.As all the angles are same so each angle must be $120$ degrees.

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Consider a line rotating about a regular $n$-gon in the Euclidean plane. If the angle at each vertex is $a$, then the line rotates $180-a$ at each vertex.

Since there are $n$ vertices, the total rotation is $n(180-a)$.

But this total rotation is $360$. So, $n(180-a) = 360$ or $a = 180 - 360/n$.

For $n = 6$, $a = 180-360/6 = 180-60 = 120$.

Note: This is not original by me.

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