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.
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.
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 ).
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 .
- 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.
- 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.)