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.)