A regular hexagon is inscribed in a circle of radius $1$. Find the measures of the segments $AB$, and $AC$. Show triangle $ABC$ is a right triangle. A regular hexagon is inscribed in a circle of radius $1$. Find the measures of the segments $AB$, and $AC$. Show that $ \bigtriangleup ABC$ is a right triangle.

Hi! I need help with this one. I was thinking of drawing a segment from $B$ to $C$ would give me the diameter of the circle so then $BC=2$ since the radius is $1$. Then because the radius is $1$, if from the center of the circle I draw two segment, one to $A$ and the other to $B$ would be an equilateral triangle and then $AB=1$. If I'm right then with Pythagoras Theorem I find $AB$. Let me know if what I'm thinking makes sense.
How do I show is a right triangle?
Thanks.
 A: Let be the point $O$ the center of the circle and $\overline{\rm AB}$ the diameter of the circle. To show that the $ \bigtriangleup ABC$ is a right triangle, we have to show that the $\angle BAC$ is a right triangle. Then we have that the $\angle BOC$ is a straight angle, so its measure is $180^{\circ}$.

$$m\angle BOC=2\cdot m\angle BAC$$
$$180^{\circ}=2\cdot m\angle BAC$$
$$\frac{180^{\circ}}{2}=\frac{2}{2} \cdot m\angle BAC$$
$$90^{\circ}=m\angle BAC$$
Therefore the $m\angle BAC$ is a right triangle and therefore $\bigtriangleup ABC$ is a right triangle.
Then, if we draw a segment from point $O$ to point $A$ and from point $O$ to point $C$ is formed $\bigtriangleup OAC$ and is gonna be an equilateral triangle. Because $\overline{\rm OA}$ and $\overline{\rm OC}$ are also radius of the circle then $OA=1$ and $OC=1$. Therefore, $AC=1$.

Last,because $\bigtriangleup ABC$ is a right triangle the Pythagoras Theorem is satisfied and we can use it to find $AB$. If the radius of the circle is $1$ then the diameter $\overline{\rm BC}$ measures $2$.

$$b^2+c^2=a^2$$
$$1^2+c^2=2^2$$
$$1+c^2=4$$
$$c^2=4-1$$
$$c^2=3$$
$$\sqrt{c^2}=\sqrt{3}$$
$$c=\sqrt{3}$$
Therefore, $AB=\sqrt{3}$
