Cyclic quadrilateral of any rectangle

Question

Is there proof that any rectangle is a cyclic quadrilateral?

Context: in Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic.

Answer 1

Hint:-

Try to solve it from the converse.Imagine a rectangle $ABCD$.Now,draw diagonal $BD$.Now,separate out triangle $ABD$.Taking BD as the diameter draw a circle which will pass through $A$ (why?).Now,if you take any arbitrary point $C$ on arc $BCD$,$\angle BCD=90^0$(why?).

Now,you can understand how to relate $90^0$ with the problem.Do the same by drawing the other diagonal such that the one of the two pairs of the angles formed is $90^0$.The other angle automatically becomes $90^0$(why?).

So,now yo get a figure with all angles $90^0$ which is a rectangle.Saying simply,a cyclic quadrilateral with one pair of adjacent angles $90^0$ is a rectangle (or any cyclic parallelogram is a rectangle).

Note-This is more or less like the converse.Hope this helps.

Answer 2

Why not just use the property of rectangles that diagonals of rectangle bisect each other?

That will mean that the intersection of diagonals is a point from which all the vertices are equidistant, which means, from the definition of a circle, that they lie on the circle with the center at the intersection.

If you had to prove the property first, you can do it by constructing the two diagonals, intersecting at O, then showing that any two vertically opposite triangles are congruent.