Question 1113755
<pre>
For a parallelogram to be inscribable in a circle, its diagonals must
be equal in length and also bisect each other.

This is true of any rectangle:

{{{drawing(400,400,-5.4,5.4,-5.4,5.4,circle(0,0,5),
line(-4.5,-2.179449472,4.5,-2.179449472), line(4.5,-2.179449472,4.5,2.179449472), line(4.5,2.179449472,-4.5,2.179449472), line(-4.5,2.179449472,-4.5,-2.179449472),
green(line(-4.5,-2.179449472,4.5,2.179449472),line(-4.5,2.179449472,4.5,-2.179449472))  )}}}


The two green diagonals are equal in length and bisect each other,
forming 4 radii of a circle.

The diagonals of a parallelogram bisect each other but are not equal unless
the parallelogram happens to be a rectangle.

Therefore a parallelogram cannot be inscribed in a circle unless the
parallelogram happens to be a rectangle.

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Since a square is a rectangle, any square can be inscribed in a circle.

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The diagonals of a rhombus bisect each other but are not equal unless
the rhombus happens to be a square.

Therefore a rhombus cannot be inscribed in a circle unless the
rhombus happens to be a square.

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One case in which a kite can be inscribed in a circle is when the
kite happens to be a square.  But that is not the only case:

{{{drawing(400,400,-1.3,1.3,-1.3,1.3,circle(0,0,1),

line(-cos(pi/6),sin(pi/6),0,1),line(cos(pi/6),sin(pi/6),0,1),
line(-cos(pi/6),sin(pi/6),0,-1),line(cos(pi/6),sin(pi/6),0,-1),
green(line(0,-1,0,1)))

 )}}} 

In order for a kite to be inscribable in a circle, the two congruent
triangles which the long diagonal divides the kite into must be right 
triangles.  That's because the only triangles inscribable in semicircles 
are right triangles.

Therefore a kite cannot be inscribed in a circle unless it contains
two opposite internal right angles.  

Edwin</pre>