Apply the Pythagorean theorem to triangles and :
Assume that in the diagram above one has . As before, , giving
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:
From the previous example we had and , so:
From the diagram
Note that (example 4) and , where is the circumradius.
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.
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).
where and .
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.
- (Long chain) For any obtuse triangle , PROVE that each of the following statements implies the others:
- 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.
- Consider with side-lengths and altitudes .