*right triangle*with an altitude as shown below:

the geometric mean theorem states that

(1)

As shown here, equation (1) is equivalent to the Pythagorean identity:

(2)

However, the equivalence holds because the altitude is *internal*. In the case of an *external* altitude, we present an analogous geometric mean theorem:

(3)

Here, equation (3) is equivalent to:

(4)

We have

## External altitudes

In examples 2 and 3 below, we prove the equivalence between equations (3) and (4).

*geometric mean theorem*for obtuse triangles.

Apply the Pythagorean theorem to triangles and :

Assume that in the diagram above one has . As before, , giving

## Extra attributes

When a triangle satisfies equation (4), it is peculiar in its own right, but also shares some things in common with the right triangle.

Note that (by definition) and , where and are the lengths of the altitudes from and . From the above diagram:

Similarly, .

From the previous example we had and , so:

From the diagram

Note that (example 4) and , where is the circumradius.

And

(Along the line we used equation (4) and its equivalent representation .)

Conclusion: or .

The relation also holds for right triangles. Using equation (4) and the cosine formula, we obtain

This is probably the easiest way to see the relationship between right triangles and triangles satisfying equation (4).

From example 7 above we had . But then also, giving .

Our triangle is obtuse so we choose .

Using , the cosine formula, and skipping a couple of steps, we have:

The second possibility is just equation (4) re-arranged.

## Essential awareness

The preceding examples show that our newest arrival — any satisfying equation (4) –“rivals” the right triangle in many aspects. More on that later. For now, any triangle whose side-slopes form a geometric progression is also comparable to a right triangle. There, a *modified* geometric mean theorem holds (see example 10).

(5)

where and .

## Earlier aim

Our original plan was to focus on equation (5) throughout today, but we were redirected along the way. Kindly bear with us for not completely fulfilling the promise made in example 8 of our post on December 28, 2020.

The foot of the altitude from is , as shown above. By the distance formula:

The slopes of the sides of form a geometric progression with and , so we can now alulate :

Finally we compare both and :

Equal. Equation (5) is satisfied.

## Takeaway

For any *obtuse* triangle , the following statements are *equivalent*:

- the orthic triangle is isosceles
- the geometric mean theorem holds.

We have a longer chain of equivalence in the exercises below.

## Tasks

- (Long chain) For any
*obtuse*triangle , PROVE that*each of the following statements implies the others*:- or
- the orthic triangle is isosceles
- the geometric mean theorem holds.

(Some of the statements above are satisfied in a*right*triangle, but the entire chain of statements are__no longer equivalent__in that case.) For something different, some of the above statements are clearly redundant; notwithstanding, just admire the sheer quantity. The aim is to increase the numbers until as many statements as there are in the Invertible Matrix Theorem are reached (and “breached”).

- (Linear combination) Suppose that satisfies equation (4).
- PROVE that .
- Write in the form .

- Consider with vertices at , , . Denote its orthocenter and circumcenter by and as usual.
- Verify that . Hence, deduce that (in other words, equation (4) is satisfied)
- PROVE that . Hence, deduce that radius is parallel to side
- PROVE that . Deduce that the orthocenter is the reflection of vertex over side .

- PROVE that the two statements below are
*equivalent*for a triangle :- it is a
*right*triangle - its area is , the product of the altitude and median from the
*same*vertex.

- it is a
- Consider with side-lengths and altitudes .

*right*triangle if . What name would you give to a triangle in which ?