- can be built from any non-isosceles right triangle
- contain a pair of angles facing each other
- combine some properties of right kites with some properties of quasi-harmonic quadrilaterals.

Thus, they are special.

## Partial proofs

Throughout we work with a non-isosceles right triangle in which . and denote the circumcenter and nine-point center.

To see a basic construction of such a cyclic quadrilateral, take any non-isosceles *right* triangle in which . Denote its circumcenter by , its nine-point center by , and its circumradius by . Consider triangles and with circumcenters and . Let the segment intersect the altitude from at point . Then the quadrilateral , with the foot of the altitude from , is cyclic.

Consider the convex quadrilateral in the diagram below:

The interior angles at and are both , while the interior angles at and are and .

*non-isosceles*.

If the parent triangle is isosceles, the altitude coincides with the median , in which case is just a line.

Indeed, , and both are equal to , where is the circumradius of the parent triangle .

Since as stated, the nine-point center is the midpoint of segment . Triangles and both contain the segment . Thus are co-linear because they all lie on the right bisector of .

## Takeaway

Let be an *acute* triangle with circumcenter and circumradius . Further, let and be the diameters of the circumcircles of and . Then the following statements are *equivalent*:

- is a right triangle in which
- the identity holds.

In case the parent triangle is *obtuse*, we have the following analogous *equivalence*:

- the side-lengths satisfy
- the identity holds.

## Task

- (Inverted squares) Let be a right triangle in which , the circumcenter is , and the circumradius is . PROVE that:
- the diameters and of the circles and are related to via an inverted Pythagorean identity:

(1)

(The relation in (1) also holds when the parent right triangle is isosceles, but the quadrilateral degenerates in that case. More on this later.)

- the diameters and of the circles and are related to via an inverted Pythagorean identity:
- (Impossible situation) Let be a right triangle in which , and let and be as in the discussion.
- Find a condition under which a circle centred at will be externally tangent to the circle as in the diagram
- PROVE that if the circumcircles of and touch externally, then .
- Deduce that it is
*impossible*for the circumcircles of and to touch externally.