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)

*non-right*, let be the foot of the altitude from vertex , and let (or ). If the side-lengths satisfy equation (1), PROVE that .

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 .