I need a theoretical proof that each angle of a regular hexagon is $120^\circ$.
[Math] Prove that the each angle of regular hexagon is 120.
geometry
Related Solutions
Suppose we construct an alternative point $P'$ in a slightly different way. Draw the line through $C$ of the hexagon such that $\angle DCG = 50$, same as in your diagram, and ignore all the rest. Then draw point $P'$ on that line such that $BC=BP'$. That gives this picture:
Using the fact that $AB=BC=BP'$ you now have isosceles triangles $BCP'$ and $BP'A$, and you should have no trouble finding all the angles in those two triangles.
Happily it turns out that $\angle BAP' = 50$, so that the point $P'$ is actually the same as point $P$, and $\angle APB = \angle AP'B$.
Let $I$ be a center of hexagon. Then $HG = ID$ and they are parallel, so $IDGH$ is a paralelogram so $K$ is also the midpoint of $GI$, thus $G,K,I$ are collinear.
Since $GEI$ is isosceles triangle and $\angle GEI = 150^{\circ}$ we have $\theta = 15^{\circ}$.
Best Answer
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
$$6(180^\circ-2x)=360^\circ\Longrightarrow 180^\circ-2x=60^\circ\Longrightarrow 2x=120^\circ$$