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## A special cyclic quadrilateral II

Following our previous discussion, we now consider quadrilaterals:

• where a pair of interior opposite angles are both
• with one side and one diagonal having equal lengths.

For today’s purpose, let’s give the pseudonym special to cyclic quadrilaterals having the above two properties, and then prove that in any such special quadrilateral:

• there’s a side whose length is twice the distance between the midpoints of the diagonals
• the diagonals and the distance between the midpoints of the diagonals are related via

(1)

## Required knowledge

Euler’s quadrilateral formula is the main background needed in order to follow today’s discussion. It’s basically the equation

(2)

connecting the side-lengths of a quadrilateral with the diagonals and the distance between the midpoints of the diagonals.

## Right kites

In a right kite, equation (2) simplifies to equation (1), that is , as shown in our first example below.

Let be a right kite with longer diagonal of length and shorter diagonal of length . If is the distance between the midpoints of the diagonals, PROVE that .

Consider the right kite shown below, with longer diagonal , and shorter diagonal .

Since in the kite shown above and because it’s a right kite, Euler’s quadrilateral formula (2) gives

Let a cyclic quadrilateral with side-lengths be special, in the sense of the two properties at the introduction. PROVE that .

This example shows that apart from right kites, there are other (cyclic) quadrilaterals that satisfy equation (1), including the ones with the pseudonym special, which is our focus for today.

Such a special quadrilateral contains two interior opposite angles measuring ; place them at and .

If we let diagonal , then , and so by Euler’s quadrilateral formula (2):

Let a cyclic quadrilateral with side-lengths be special, in the sense of the two properties at the introduction. PROVE that there is a side whose length is twice the distance between the midpoints of the diagonals.

Using the same diagram/notation for the previous example, we have:

Since one of the diagonals has the same length as one of the sides, it can’t be diagonal because is the diameter. So let’s suppose diagonal as the same length as side . Then the above equation gives .

## Rare kind

Consider a right kite in which the interior angles at and are and . Show that is special in today’s terms.

Diagonal will now have the same length as sides and .

Show that the right kite in example 4 can be obtained from any isosceles triangle .

Suppose that . Let and be the circumcenter and orthocenter of the parent triangle . Then the quadrilateral is a particularly pleasant right kite.

## Takeaway

Let be a quadrilateral with side-lengths , diagonals , and the distance between the midpoints of the diagonals. If the interior angles at and are both , then the following statements are equivalent:

1. .

• (Special property) Let be special, as per the definition at the introduction; specifically, let the interior opposite angles at and be both , and let diagonal have the same length as side . PROVE that:
1. the radius through is parallel to side
2. the orthocenter of triangle is a reflection of vertex over side .

## 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.