True? "A rectangle is a subset of a circle if and only if all its vertices are in the circle (or its boundary)." Is this statement true?

A rectangle is a subset of a circle if and only if all its vertices are in the circle (or in its boundary).

Intuitively it's true, but can someone give me proof?
Thank you in advance!
 A: The statement is true, but I don't know of an elementary proof.  The proof that comes to mind uses the Krein-Milman Theorem: a compact set is equal to the closed convex hull of its extreme points.
Suppose all the vertices of a rectangle are inside the circle.  Then every convex combination of the vertices is also inside the circle (since the circle is also convex).  Since the set of all convex combinations of the vertices generates the rectangle, the rectangle is inside the circle.
A: Your statement is true. A proof relies on the fact that the points in a circle form a convex set. Suppose the four vertices of the rectangle are in the circle. Then because the circle is convex, the four edges of rectangle must also be in the circle. Then if you take any point on the interior of the rectangle, it must lie on a line-segment connecting a vertex of the rectangle to some point on an edge of the rectangle, so it too is in the circle.

Just a note on terminology, many people use the words circle and rectangle to refer to just the boundary of those shapes. 
A: Circle is convex. Thus, for any 2 points in a circle, there is a straight line entirely contained in a circle that connects our 2 points. Thus, if 4 vertices of a rectangle are in a circle, so are the boundaries of a rectangle. Can you think about an argument explaining why every interior point of a rectangle is in your circle as well?
