Let be the side-lengths of
, and let
be the length of the bisector of
, as shown below:
Then is the geometric mean of
and
if, and only if:
We’ll see the simplifications that result when equation (1) restricts to right triangles
and the simplifications that result when (1) is considered in the context of triangles satisfying
Let’s set the ball rolling with what a right triangle enjoys solely.
Suppose that is a right triangle with
. Then the median from vertex
is half of the length of the hypotenuse:
and so
is the geometric mean of
and
. On the other hand, if a triangle has the property that one of its medians is the geometric mean of the two segments on the opposite side, for example,
, we show that the triangle is a right triangle. Indeed:
So the triangle is a right triangle. Since all medians are internal, this property is unique to right triangles.
Formula derivation
Our main result is example 4, for which we may need the next two examples.




We first have that . Also:
The sine rule applied to s
and
then gives:
Basically, . This is very useful in finding the coordinates of the foot of an angle bisector, which in turn is useful in finding the coordinates of the incenter of a triangle.



Let’s refer to the diagram in example 2 again:
and then apply Stewart’s theorem to to obtain
We now use the fact that and
from example 2.
An alternative procedure is to use area arguments as well as the cosine formula and double angle identities for and
. This way one avoids going through Stewart’s theorem and an explicit determination of
and
.





First suppose that is the geometric mean of
and
. Then
.
Conversely, suppose that .
Notice how we obtain another equivalence: . Translate into words: we killed two birds with one stone.




The foot of the bisector of
has coordinates
. By the distance formula:
Also:
Thus, is the geometric mean of
and
.
Fifteen degrees
We now restrict equation (1) to right triangles and to triangles satisfying equation (3) . It turns out that the interior angles are all multiples of .

Using the fact that and
, we have:
The angles of the triangle will then be ,
,
. All multiples of
.

In this case . Together with equation (1), we have
Alternatively, the bisector is a median in this case, and so the conclusion follows from example 1 and example 4.



We have: . Isolate
and simplify:





Since (see here), we have (by example 7 above):
Also, and
yield
and
(or
and
).
Further developments
Here are two out of a few more consequences.





We have and
. Eliminate
to obtain
.







In example 4 we had ; in example 9 above we had
. Now let’s calculate
:
We see that form a geometric progression with common ratio
.
Takeaway
In triangle , let
be the foot of the bisector of
. Then the two statements below are equivalent:
is the geometric mean of
and
is the geometric mean of
and
.
Further, for any triangle whose side-lengths
satisfy equation (1), the following two statements are equivalent:
is a right triangle
.
Finally, for any triangle whose side-lengths
satisfy equation (1), the two statements below are equivalent:
is a pseudo right triangle
.
Tasks
- Using
with vertices at
,
,
, verify that:
- its incenter is
.
- the ratio of the coordinates
of the incenter is golden; that is, it satisfies
.
- the slope of the bisector of
is
, the negative of the golden ratio.
- its incenter is
- Let
be the side-lengths of triangle
.
- PROVE that the harmonic mean of
and
is
.
- Deduce that the bisector of
is the harmonic mean of the two segments on the opposite, if
.
- PROVE that the harmonic mean of
- Suppose that the side-lengths
of triangle
satisfy equation (3), namely
.
- PROVE that the length
of the bisector of
satisfies
.
- Deduce that
is the harmonic mean of
and
.
- PROVE that the length
- Suppose that
satisfies the usual Pythagorean identity
.
- If the bisector of
satisfies equation (1) adapted as
, PROVE that
.
- Under the above, deduce that
,
.
- If the bisector of
- In
, let the side-lengths be
, and let
denote the circumradius, altitude from
, and the length of the bisector of
. If
satisfies equation (3), PROVE that the squared lengths
form a geometric progression with common ratio
.
- In
, let
denote the circumradius, altitude from
, and the median from
. If
satisfies equations (1) and (3), PROVE that
.
(Basically a right triangle can be formed with the medianas the hypotenuse.)