*mirror*the number of statements in the Invertible Matrix Theorem, to

*mimic*right triangles, as well as to have some

*memory*of , we’ve assembled statements that are equivalent in certain triangles. Most of these statements are trivial, but what is vital is that right triangles and the triangles that satisfy the statements are rivals. Vital rivals.

## Problem statement

*non-right*triangle with side-lengths , altitudes , circumradius , circumcenter , orthocenter , and nine-point center . Then the following statements are

*equivalent*:

- the orthic triangle is obtuse isosceles
- radius is parallel to side
- the nine-point center lies on
- the geometric mean theorem holds
- the bisector of has length , where
- segment is
*tangent*to the circumcircle at point - the orthocenter is a reflection of vertex over side .

The connection between the orthic triangle and the geometric mean theorem seems cool. The restriction to *non-right* triangles is crucal: some of the statements hold in right triangles, but not the entire chain of statements.

Our favorites: . Notice the last two.

## Partial solutions

A complete proof will require about implications, and we’ve already seen some of these implications in previous iterations of our discussions: May 14, May 28, June 14, and June 28. Below we make some random selections.

*parallel*to side .

To be specific, let and set . Then and . Consider the circumcircle shown below:

The angle which the *major arc* subtends at the center of the circle is *twice* the angle it subtends at the circumference, and so:

Since is isosceles, we have that

This shows that is parallel to .

Begin by drawing the circumcircle:

If , then , as marked above. Reason: radius is parallel to side by assumption. Next, is isosceles, and so . In turn, this yields . The major arc now subtends an angle of at the center of the circle. This means that it subtends an angle of at the circumference. Thus, . The difference between and is then .

A different orientation having angle placed at will also yield the same conclusion.

Let and set . Then and . Draw the circumcircle, like so:

Since is *obtuse*, its orthocenter is situated outside the triangle as shown above. Join and . Extend side to meet at . Since now becomes the foot of the altitude from , we have that . Since is parallel to , it follows that .

Radius is perpendicular to tangent at the point of contact. Implies: is a tangent to the circumcircle at .

Apply the Pythagorean theorem to the right triangle below:

We had from the preceding example. This in turn implies . So below is isosceles with :

Extend side to meet at . Since now becomes the foot of the altitude from , we have that . Altitude bisects the base, so . This proves that is a reflection of over side .

We use the *segment addition postulate*. Note that in any triangle with orthocenter , circumcenter , and nine-point center we have:

We proved before that when . So

After some simplifications, we obtain

Similarly:

Let’s examine the absolute values. There are four cases to consider.

First, we can’t have and simultaneously. Otherwise, their sum must be greater than zero as well; but their sum is .

Next, suppose that and . Then the sum is , and so:

Because , this leads to . This is a special case. If the points aren’t co-linear, then in , the median through passes through the nine-point circle, and so the length of this median is a radius of the nine-point circle, namely . We now have a triangle in which the sum of two sides is and a median has length . This is impossible (see the exercises at the end). Indeed, the side-lengths of have to be of the form for sides (or sides ). If we compute the cosine of the angle at , we obtain

The third and fourth cases are the same. For example, and . Then take

and obtain

Using the fact that , we have:

After all the trouble of the previous example, we don’t want to bother you with another seemingly lengthy procedure, but note that results after simplifications.

Normally, . So

First isolate from as and then use in the standard angle bisector formula:

## Peculiar scenario

Below is a reward of the labour in example 6.

Take with . Then . The nine-point center is precisely , since (according to example 6):

Our next post will show that no other triangle has this property.

## Takeaway

Let be a *non-right* triangle with side-lengths , altitudes , circumradius , circumcenter , orthocenter , and nine-point center . Then the following statements are *equivalent*:

- radius is parallel to side
- segment is tangent to the circumcircle at point
- the nine-point center lies on
- the orthic triangle is obtuse isosceles
- the geometric mean theorem holds
- the bisector of has length , where
- the orthocenter is a reflection of vertex over side .

No need to wait until 2026 for *twenty six* equivalent statements.

## Tasks

- (Identical traits) Consider a
*right triangle*with side-lengths , circumradius , circumcenter , nine-point center , and . Let be the circumcenter of and let be its circumradius. Similarly, let be the circumcenter of and let be its circumradius.- Show that are co-linear.
- PROVE that and .
- Deduce that the ratio in which the bisector of divides side is the same as the circumradii ratio .

- (Identical traits) Consider a
*non-right*triangle with side-lengths , circumradius , circumcenter , and nine-point center . Let be the circumcenter of and let be its circumradius. Similarly, let be the circumcenter of and let be its circumradius. If :- Show that are co-linear.
- PROVE that and .
- Deduce that the ratio in which the bisector of divides side is the same as the circumradii ratio .

(There goes a non-right triangle that*mimics*a right triangle in many aspects.)

- (Isosceles trapezium) Suppose that an obtuse satisfies any of the equivalent statements considered in this post. Let be a point on the circumcircle such that is a diameter.
- PROVE that .
- Deduce that quadrilateral is an
*isosceles trapezium*.

- (Impossible triangle) In triangle , suppose that and that the length of the median from is , as per the special case encountered in the course of example 6.
- PROVE that the lengths of sides and must be and (or the other way).
- Using the triangle inequality, deduce that the three points are co-linear.

- PROVE that the following two statements are
*equivalent*for any triangle :- the length of the bisector of satisfies .