🚀 go-pugleaf

In article <5hs7ff$4qd@opus.vcn.bc.ca>
          landmark@vcn.bc.ca (Scott Phung) wrote:

->margot@cnwl.igs.net (David) wrote:
->D> landmark@vcn.bc.ca (Scott Phung) wrote:
->D>
->D>
->D> Scott,  a cyclic quadrilateral is formed as *any* four distinct points
->D> on the circle.  The area is not fixed in any way, and the center of
->D> the circle may be far from the intersection of the diagonals.  You can
->D> find the area generally only as the area of the quadrilateral.  Being
->D> cyclic adds no distinct differences to that figure so far as area is
->D> concerned, although there is a wealth of other properties, such as "A
->D> cyclic quadrilateral that is a parallelogram must be a rectangle."
->D> ..in which case the center of the circle *would* coincide with the
->D> intersection of diagonals.
->D>
->D> David.
->D> If knowledge is power, that explains my weakness.  :-)
->
->David,
->
->The vertices can be on *any* four distinct points?
->But then does the diagonaly opposite angles add up to 180?
->
->If it does, my mistake, and I will beleive that what your saying
->is correct.
->
->Or maybe becuase the vertices lie on the circle makes the
->diagonally opposite angles add to 180.

This works both ways, and is proved in Euclid.

Choose any four points on the circumference of a circle and join them
to form a quadrilateral. Then opposite angles are supplementary.

Draw any quadrilateral with opposite angles supplementary. Then a
circle exists which passes through all four points.

Hope that helps.

-- 
Malcolm in sunny Berkhamsted