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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. .