![Rendered by QuickLaTeX.com h_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-12f43127818fc8246bd1d6642e18ee7c_l3.png)
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)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(-6,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79c16eb208dc820695dfd9a30fa1c847_l3.png)
![Rendered by QuickLaTeX.com B(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7045877de833cb7e53ba1ef4ddfaeb24_l3.png)
![Rendered by QuickLaTeX.com C(2,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e2fb9eee47c72cdf6ab23e648c29df8c_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com h_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-12f43127818fc8246bd1d6642e18ee7c_l3.png)
![Rendered by QuickLaTeX.com h_C=\sqrt{AF_C\times BF_C}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b85efafd7be78cc2b773d4b53d8984fa_l3.png)
We have
External altitudes
In examples 2 and 3 below, we prove the equivalence between equations (3) and (4).
Apply the Pythagorean theorem to triangles and
:
![Rendered by QuickLaTeX.com \triangle ACF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ba3bfa405a0aab02705feceec7367361_l3.png)
![Rendered by QuickLaTeX.com \triangle BCF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-84a9ecb587d3f133d97d8d6670321e64_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com AF_B=AF_A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1a9537dcf70d0094b2b9aaf373504698_l3.png)
![Rendered by QuickLaTeX.com BF_A=BF_B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-46fda51e0095f156b4c5505aee92afe8_l3.png)
![Rendered by QuickLaTeX.com F_A,F_B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b864299f8df09b4d785f0a52be4fb558_l3.png)
![Rendered by QuickLaTeX.com A,B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-99d471de5f012da92e07460b8403a0f1_l3.png)
Note that (by definition) and
, where
and
are the lengths of the altitudes from
and
. From the above diagram:
Similarly, .
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com ap=cs](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7c2693acd698a3330bf7e03809cff7_l3.png)
From the previous example we had and
, so:
From the diagram
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com p(p+a)=qr](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6042a974f48a6578d32c984a913017ee_l3.png)
Note that (example 4) and
, where
is the circumradius.
And
(Along the line we used equation (4) and its equivalent representation .)
Conclusion: or
.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com \cos^2A+\cos^2B=1](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1f72bb6b73b10843e09278914419cc25_l3.png)
The relation also holds for right triangles. Using equation (4) and the cosine formula, we obtain
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A-B=90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b329938a567965f4d8a4958871a5e3f1_l3.png)
![Rendered by QuickLaTeX.com B-A=90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-33051db39fec2563ff38b91d6d9f2da1_l3.png)
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 .
![Rendered by QuickLaTeX.com \cos^2A+\cos^2B=1](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1f72bb6b73b10843e09278914419cc25_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
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).
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com AB,BC,CA](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4950f2601ab65c5b90b958ee89e0e4c4_l3.png)
![Rendered by QuickLaTeX.com a,ar,ar^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-05d26acad480937e0cbaccc666187197_l3.png)
![Rendered by QuickLaTeX.com h_A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0fde55553cab4b72e169b87b146291bc_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
(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.
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com A(1,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-671ce02f23cbb90f33dabfb6ed108046_l3.png)
![Rendered by QuickLaTeX.com B(3,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b67631b2dbdedc7cc3d349be4e9a2d8_l3.png)
![Rendered by QuickLaTeX.com C(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-36581d8d833c92827ae6e6709908f73d_l3.png)
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
al
ulate
:
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
.
- PROVE that
- 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
.
- Verify that
- 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.
- Consider
with side-lengths
and altitudes
.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A+B=90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ebe1488b6253ce9a46404a6c936c64bf_l3.png)
![Rendered by QuickLaTeX.com A-B=90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b329938a567965f4d8a4958871a5e3f1_l3.png)