SOLUTION: How to prove the statement " If a parallelogram has a circumscribed circle, then it is a rectangle " and its converse " If a parallelogram is a rectangle, then it has a circumscrib

Let a parallelogram ABCD has a circumscribed circle.

Then, since in a parallelogram the opposite sides are congruent, these arcs are congruent in pairs:

   arc(AB) ~ arc(CD)   and   arc(BC) ~ arc(DA)   ("~" means "congruent arcs").

Then the arcs  arc(AB) + arc(BC)  and  arc(CD) + arc(DA) are congruent:

    arc(AB) + arc(BC)   ~   arc(cd) + arc(DA).


From the other side, their sum 

    (arc(AB) + arc(BC))   +   (arc(CD) + arc(DA))

is the entire circle, i.e 360°.


It implies that  

arc(AB) + arc(BC) = 180°  and 

arc(CD) + arc(DA) = 180°.


Then each of the two angles ABC and CDA is the right angle, since each of the angles leans the 180° arc.


Thus we proved that the given parallelogram has at least one (actually, two) right angles.
Hence, the parallelogram is a rectangle.

This statement is easier to prove.


a) Since in any rectangle the diagonals are congruent, and
   since in any parallelogram the diagonals bisect each other, 
   it implies that the intersection point of diagonals of the rectangle is equidistant from its verices.


   Thus the intersection of diagonals is the center of the circle circumscribed about the rectangle.