- ,
- ,
- ,

What makes them special is the fact that their side-lengths satisfy the modified Pythagorean identity

(1)

with the accompanying equivalent descriptions. The first triangle has already appeared on August 14 and on October 14. The second triangle is making its debut here.

Since , the conclusion follows from one of the equivalent statements here.

In example 1 above we saw that the side-lengths satisfy equation (1). As such , by one of the equivalent statements here. So:

A direct computation of the side-lengths and can also be used to prove the above relationship, but that’s a longer procedure.

By example 2 above we had . Re-write as . Using one of the equivalent statements here, we know that .

One can also prove this by a direct computation of the circumradius and the fact that . Definitely a longer procedure.

Let’s use the cosine formula and the fact that :

To show that , re-write as and then use the fact that just proved.

Take the tiny triangle where the vertices are located at , , and .

## Takeaway

Let be the side-lengths of , and let be the radius of its circumscribed circle. If equation (1) holds, then the following statements are *equivalent*:

- .

Note the first one.

## Task

- (Late fifties) 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, the orthocenter, the midpoint of side , the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following*fifty-eight*statements are*equivalent*:- or
- 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
- 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
- segment is tangent to the nine-point circle at .

( short of the target.)

- (Extra feature) If satisfies equation (??), PROVE that its nine-point center divides in the ratio .