- 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:
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.





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



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):


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






Diagonal will now have the same length as sides
and
.


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:
.
Task
- (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:
- the radius through
is parallel to side
- the orthocenter of triangle
is a reflection of vertex
over side
.
- the radius through