This is a paragraph.

# A special cyclic quadrilateral I

The class of cyclic quadrilaterals to be considered in this (and possibly next) post:

• 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.

Explain how to construct a cyclic quadrilateral that exhibits the properties hinted at above.

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. Suppose that the segment intersects the altitude through at point . PROVE that is a cyclic quadrilateral.

Consider the convex quadrilateral in the diagram below: The interior angles at and are both , while the interior angles at and are and .

In the cyclic quadrilateral above, explain why the parent right triangle should be non-isosceles.

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

In the cyclic quadrilateral above, PROVE that one side and one diagonal are equal.

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

Let and denote the circumcenters of triangles and . PROVE that are co-linear.

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:

1. is a right triangle in which 2. the identity holds.

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

1. the side-lengths satisfy 2. the identity holds.

• (Inverted squares) Let be a right triangle in which , the circumcenter is , and the circumradius is . PROVE that:
1. 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.)

• (Impossible situation) Let be a right triangle in which , and let and be as in the discussion.
1. Find a condition under which a circle centred at will be externally tangent to the circle as in the diagram 2. PROVE that if the circumcircles of and touch externally, then .
3. Deduce that it is impossible for the circumcircles of and to touch externally.

Posted on