Writing for the third time on the geometric mean theorem, after an external eye scrutinized and stamped the previous two writings on the same theme, seems a joy thing.

With that said, today’s post focuses on numerical problems that concretize the geometric mean theorem in *obtuse* triangles.

Wow! Thanks for clicking.

Note that the triangle in this first example is a 30-30-120 triangle; it satisfies the geometric mean theorem *twice*.

From the above diagram, . Also and . Thus:

Similarly, if denotes the foot of the altitude from vertex , then:

Another unique property of the triangle.

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

## Takeaway

The two statements below, the geometric mean theorem, and (more than) 75 other statements, are *equivalent* in any non-right triangle with circumradius and side-lenths :

- .

## Task

- (Aufbau) In triangle , let be the side-lengths, the circumradius, the circumcenter, the nine-point center, and the orthocenter. PROVE that the following statements are
*equivalent*:- the reflection of over is
- the circle with diameter passes through vertices and
- radius is parallel to side and radius is parallel to side
- the reflection of over is and the reflection of over is
- is the geometric mean of and , and is the geometric mean of and . ( and are the feet of the altitudes from and , respectively.)