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
- 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 .

- 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 .

- Suppose that satisfies the usual Pythagorean identity .
- If the bisector of satisfies equation (1) adapted as , PROVE that .
- Under the above, deduce that , .

- 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 median as the hypotenuse.)