🚀 go-pugleaf

In article <5hooqn$ikg$1@nntp.igs.net>, margot@cnwl.igs.net (David) wrote:

> landmark@vcn.bc.ca (Scott Phung) wrote:
> 
> 
> Scott,  a cyclic quadrilateral is formed as *any* four distinct points
> on the circle.  The area is not fixed in any way, and the center of
> the circle may be far from the intersection of the diagonals.  You can
> find the area generally only as the area of the quadrilateral.  Being
> cyclic adds no distinct differences to that figure so far as area is
> concerned, although there is a wealth of other properties, such as "A
> cyclic quadrilateral that is a parallelogram must be a rectangle."
> ..in which case the center of the circle *would* coincide with the
> intersection of diagonals.
> 
> David.
> If knowledge is power, that explains my weakness.  :-)

I think that the area of a cyclic quadrilateral is equal to sqrt
((s-a)(s-b)(s-c)(s-d)), where s is the half-circumference of the
quadrilateral. 

meeroh

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