If a non-right triangle, like the one shown above, satisfies the identity
(1)
then we can recast equation (1) in a form that looks more like the usual Pythagorean identity, namely:
(2)
As , we have an approximate Pythagorean identity
(3)







Consider the following diagram:
Let be the circumradius and suppose that
(if
, take
). We’ll use some properties that are consequences of equation (1), including the one from our previous post:
Then:








From the diagram below
we get . Equation (1) holds:
as well as equation (2), because:








Note that and
can switch places. From the diagram below
we get . Equation (1) holds:
as well as equation (2), because:








From the diagram below
we get . Notice that equation (1) holds:
as well as equation (2), since:








From the diagram below
we get . Equation (1) holds:
as well as equation (2), since
As in this post, notice that the title of today’s post can be abbreviated as API, an absolutely endearing term in computer programming.
Takeaway
In a non-right , let
be the side-lengths,
the circumradius,
the circumcenter,
the orthocenter, and
the nine-point center. Then the following statements are equivalent:
is isosceles with
is isosceles with
is the circumcenter of
is right angled at
.
Don’t confuse with the functional group of an aldehyde.
Task
- (Early fourties) In a non-right triangle
, let
be the side-lengths,
the altitudes,
the feet of the altitudes from the respective vertices,
the circumradius,
the circumcenter,
the nine-point center, and
the orthocenter. PROVE that the following forty four statements are equivalent:
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
- radius
is parallel 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
- 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
- segment
is tangent to the nine-point circle at
.
(Wow!!! The numbers increased so rapidly, from twenty one a few days ago.)
- (Extra feature) If
satisfies equation (1), PROVE that its nine-point center
divides
in the ratio
.