For Euclidean parallelograms prove that a parallelogram has

For Euclidean parallelograms, prove that a parallelogram has a circumscribed circle if and only if it is a rectangle (Hint: for the \"only if\" part: Opposite angles must subtend sem-circles).

Solution

Necessary condition : A parallelogram has a circumscribed circle if it is a rectangle.

Now let a parallelgram is a rectangle, so clearly its both diagonals are equal to each other and bisect each other so that the distance of bisecting point from all the four vertices remains the same. That means if we consider these four equal lengths from that bisecting point ( center in this case), to be radii of a circle, that circle must surely pass from those four vertices that would lie on the circumference of the circle.

It proves the parallelogram has a circumscribed circle if it is a rectangle.

Sufficient condition (only if part) IF a parallelogram is circumscribed, then it will be a rectangle.

Now we have the property that such parallelogram will be a cyclic quad and by definition the sum of two opposite angles of a cyclic quad is always 180 degree, it is because the angle formed in the semicircle is always of 90 degree. Further we also have that a rectangle is a quad in which the sum of its two opposite angles is always 180 degree. Thus clearly that parallelogram will surely be a rectangle.

It proved the result.