Two equilateral triangles are circumscribed about the same circle such that their intersection is a regular hexagon (see picture below). If the radius of the circle is $10$, find the area and the perimeter of the hexagon.
Here's my attempt:
I know that the sum of the angles of the hexagon is $(n-2)180=4*180º=720º$. So each angle has measure $720º/6=120º$.
If I could show that angle $∠ AMO=∠ ABO=90º$, from a theorem presented previously in the book, it would follow that point $O$ lies on the bisector of $∠ A$. From there I think I can find $|AC|$ and hence the perimeter and area of the hexagon. I'm not sure how to go about showing that tough. Any help will be appreciated. Thanks.