(1)
that can be used to characterize right triangles.
To this end, let be a right triangle in which
, and
the circumcenter
the circumradius
and
the diameters of triangles
and
.
Then equation (1) holds.
Preliminary calculations










A sample diagram is shown below:
Since is the circumcenter of
, it lies somewhere on the right bisector of side
and so when
is joined to
, the segment
is then parallel to side
. Similarly,
is parallel to side
. Thus
. Now join
to
and use angle chasing to complete the similarity argument.









From the previous example, we saw the similarity between triangle and the parent triangle
:
So:
We obtain as required. Notice also that we have
(2)
Partial characterization
Partial in the sense that there’s another triangle for which the equivalence in the next two examples works. We’ll see that later.











Apply the extended sine law to triangle below, noting that
:
This would give:
Similarly, applying the extended sine law to triangle gives
Since and
, we now get:












We’ll show in a later post that the following extension of the extended sine law holds:
(3)
An absolute value may be needed in one of the cosine terms if the parent triangle is obtuse. However, since we’re working with the squares for now, there’s no need for an absolute value. We have from (3) that
(Or .) Assuming
, we then have:
Thus, either the triangle is right-angled at , or it’s an obtuse triangle that shares several properties in common with a right-triangle.
Pertinent consideration
We emphasize that the equation also holds in a certain non-right triangle setting. More on this later.




We’re to show that , where
is the circumradius,
is the diameter of the circumcircle of
, and
is the diameter of the circumcircle of
.
By calculation, the circumcenter is
. Thus the circumcenter
of
is
, while the circumcenter
of
is
. Notice that the nine-point center
shown in the diagram is co-linear with
and
, just like in a right triangle. Furthermore:
So:
Takeaway
Suppose that triangle has side-lengths
, circumcenter
, circumradius
, and
the radii of the circumcircles of
and
. Then the following statements are equivalent:
is right-angled at
.
Tasks
- (Geometric means) Suppose that the side-lengths of
satisfy
. Using the notation in this post, PROVE that:
- (Geometric means) Suppose that the side-lengths of
satisfy
. Using the notation in this post, PROVE that:
- (Growing membership) In a non-right triangle
, let
be the side-lengths,
the altitudes,
the feet of the altitudes from the respective vertices,
the midpoints of sides
in that order,
the Euler points,
the circumradius,
the circumcenter,
the nine-point center,
the orthocenter,
the reflection of
over side
,
the reflection of
over side
,
the reflection of
over side
,
the symmedian point,
the foot of the symmedian from vertex
, and
the radius of the polar circle. PROVE that the following eighty-five statements are equivalent:
or
and
divide
harmonically
is the reflection of
over side
is the reflection of
over side
is congruent to
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
is right-angled at
is right-angled at
- quadrilateral
is a rectangle
- the points
are concyclic with
as diameter
- the reflection
of
over
lies internally on
- the reflection
of
over
lies externally on
- radius
is parallel to side
is the reflection of
over side
- segment
is perpendicular to side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
- segment
is tangent to the circumcircle at point
- median
has the same length as the segment
- the bisector
of
is tangent to the nine-point circle at
is a convex kite with diagonals
and
- altitude
is tangent to the nine-point circle at
- chord
is a diameter of the nine-point circle
- segment
is tangent to the nine-point circle at
.
(More statements reserved for later. We aim to stop atstatements, if possible.)