THE TRIANGLE’S CENTERS AND THE EULER LINE

By Dario Gonzalez Martinez

IMPORTANT TRIANGLE’S CENTERS

As the most of us probably know, the triangle has important centers from which 4 are really popular among the school mathematics. They are:

The Centroid	Commonly denoted by G
The Incenter	Commonly denoted by I
The Orthocenter	Commonly denoted by H
The Circumcenter	Commonly denoted by C or O
Table 1

We will define these important centers and prove their existence. However, before stating any definition or proof, I would like to discuss an interesting theorem which will serve as fundament for our future proofs.

CEVA’S THEOREM

This theorem has its name due to Giovanni Ceva, who proved this theorem in 1678 in his work De lineis rectis. Even though, this theorem was proved before by Yusuf Al-Mu’taman ibn Hud.

First of all, let’s define what is a cevian is any line segment in a triangle with one endpoint on a triangle’s vertex and the other endpoint on the opposite side. For example, figure 1 below shows three cevians , and in two different triangles.

Figure 1(a)	Figure 1(b)

As you probably noted, the cevians are not necessary concurrent, and they are not necessary inside the triangle. However, there exist the case where the three cevians are concurrent, and that is the case in what we are interested (does not matter if they are inside or outside of the triangle).

Ceva’s theorem: Given a triangle ABC and three points M, N and P along the triangle’s sides, a necessary an sufficient condition for the cevians , and to be concurrent is that:

Figure 2

As a result that the triangle’s centers are special cases of point Q in figure 2, the Ceva’s theorem provides us with an adequate tool for proving the existence of those triangle’s centers.

Let’s prove this important theorem. Consider the figure 2 above where the cevians , and are concurrent at the point Q.

Te first step is to draw a parallel line to through the point C as follow in figure 3 below.

Figure 3

The second step is to extend the cevians and such that they intersect the parallel line to at the points X and Y, respectively (see figure 4(a) and (b) below).

Figure 4(a)	Figure 4(b)

Figure 4(a) shows that the triangles ABM and CMX are similar (yellow ones), and so are triangles ABN and CNY in figure 4(b) (green ones). Then we can write the following equalities:

and

By multiplying these two expressions, we obtain

(1)

Now, by seeing figure 5(a) and (b), we are going to consider that triangles APQ and XCQ are similar (red ones), and so are triangles BPQ and YCQ (blue ones).

Figure 5(a)	Figure 5(b)

So, we can state the following equalities:

and

From which we can conclude

(2)

Finally, by multiplying expressions (1) and (2), we obtain

Of course, we need to prove the converse case to be done (because the above relation is necessary and sufficient condition). Thus, suppose that M, N and P are points on , and , respectively, and they also satisfy:

Let Q be the intersection point between and , and let P’ be the intersection point between and . Since , and are concurrent, we will have:

Thus, we obtain

Finally, the latter expression implicates that P’ = P, then , and are concurrent.

THE CENTROID

Definition: A median in a triangle is a line that passes through a triangle’s vertex and the midpoint of the opposite side. The three median of a triangle meet at the same point called Centroid.

Figure 6(a) below shows the median , where M is midpoint of . Figure 6(b) also shows the three medians and their intersection which is the centroid G.

Figure 6(a)	Figure 6(b)

By Ceva’s theorem, we can easily prove that the three medians are concurrent at G. Given that M, N and P are midpoints, then we have:

Thus, we obtain:

Then the median are concurrent.

THE INCENTER

Definition: An angle bisector in a triangle is a line segment that extends from a triangle’s vertex to the opposite side in such a way that it bisects the angle at the vertex. The three angle bisectors of a triangle meet at the same point called Incenter.

Figure 7(a) below shows an example of an angle bisector (). Figure 7(b) also shows the three angle bisectors and their intersection which is the incenter I.

Figure 7(a)	Figure 7(b)

Before applying the Ceva’s theorem to prove that the three angle bisectors are concurrent, we should remember the angle bisector theorem, which states that a triangle’s angle bisector divides the side where it lands into segments that stand in the same ratio as the two remaining. Thus, we will have:

Therefore, we can state the following relation:

Then, the three angle bisectors are concurrent.

Important note: It is important to consider that an angle bisector is the set of all points that are equidistant from the angle’s sides. Therefore, the triangle’s angle bisectors are equidistant from the triangle’s sides, which implicate that I (incenter) is equidistant from each triangle’s sides.

Thus, if we take I as a center and the distance from I to one triangle’s side as a radius, we can construct a circle inside the triangle.

Figure 8

Where , and are angle bisectors, and , and are distances from I to the triangle’s sides , and , respectively.

THE ORTHOCENTER

Definition: The orthocenter H is the intersection point of the three triangle’s altitudes. An altitude is a segment line drawn from a triangle’s vertex and it is perpendicular to the opposite side.

Figure 9(a) shows an example of altitude represented by the segment . At the same time, figure 9(b) shows the intersection of the three triangle’s altitudes at the point H (orthocenter).

Figure 9(a)	Figure 9(b)

To prove that the three triangle’s altitudes concur to the same point, we need to observe some relation in the triangle of figure 9(b). The altitudes drawn in triangle ABC formed pairs of similar triangles as it is shown in figure 10 below:

Figure 10(a)	Figure 10(b)	Figure 10(c)

Then, we have the following similar triangles: 1) triangles AMC and BNC, 2) triangles ABN and APC, and 3) triangles ABM and BCP (because they share an angle and have a right angle). Hence, we can write the following ratios:

Thus, if we change the denominator in the Ceva’s relation and apply the above relations, we will have

Then, the triangle’s altitudes are concurrent.

THE CIRCUMCENTER

Definition: A perpendicular bisector in a triangle is a line that passes perpendicularly through the midpoint of each triangle’s sides. The three perpendicular bisectors of a triangle meet at the same point called Circumcenter.

Figure 11(a) below shows the perpendicular bisector , where M is midpoint of . Figure 10(b) also shows the three perpendicular bisectors and their intersection which is the circumcenter O.

Figure 11(a)	Figure 11(b)

Due to the perpendicular bisectors are not cevians, I left this triangle’s center at the last. We will need a different strategy to prove that the three perpendicular bisectors are concurrent lines.

Consider figure 12 below

Figure 12

Let the lines that pass through R, S and T be perpendicular bisectors, and let O be the point where the perpendicular bisector that pass through T and R points concur. Now, we need to remember the definition of a perpendicular bisector:

A perpendicular bisector of the segment is the set of all points in the plane that are equidistant from the endpoints of that segment.

Therefore, we can write the following relationship:

For transitivity, we obtain that

This implicates that O is an equidistant point from the endpoint of the segment , which means that O belong to the perpendicular bisector that passes through S. Thus, the three perpendicular bisectors concur at the point O.

Important note: Given that a perpendicular bisector is the set of all points that are equidistant from the endpoint of each triangle’s side, then we can conclude that we can draw a circle that around the triangle’s vertices.

Figure 13

Where , and are perpendicular bisectors, and , and are distances from O to the triangle’s vertices A, B and C, respectively.

EULER LINE

Euler found an interesting relationship between three of the four triangle’s centers discussed before. This relationship involves that the Circumcenter (O), the Centroid (G) and the Orthocenter (H) are always on the same line!!!!! Does not matter from which triangle you construct them.

The Euler Line proof is equivalent to solve the problem of proving that the Circumcenter (O), the Centroid (G) and the Orthocenter (H) of a triangle are collinear.

Thus, we will consider any triangle represented by the triangle ABC in the figure 14 below:

Figure 14

We should clarify some objects presented in figure 14. First, O and G are Circumcenter and Centroid, respectively, and consider the segment which connects them. Second, consider the segments , and that are an altitude, a median and a perpendicular bisector, respectively.

The first step is to extend the segment to meet the altitude at the point H as it is shown in figure 15 below:

Figure 15

Now we can state the principal claim of the proof, which could be expressed as follow:

Main claim: If we can show that the point H is indeed the orthocenter of the triangle ABC, then we will have proved that O, G and H are collinear. That is, we will have proved the existence of the Euler Line.

Since and are both perpendicular to , then they are parallel to each other. Therefore, triangles CHG and GOF are similar (yellow triangles), and given that G is centroid, we have that:

Since triangles CHG and GOF are similar, similarly we will have

The second step is to draw segment (note that this segment intersects at point P), the median (which passes through G because it is centroid) and the perpendicular bisector (which passes through O because it is circumcenter). These new constructions are shown in figure 16 below:

Figure 16

Due to G is centroid, we have

Also, we already know that

Therefore, triangles AHG and GOM (green triangles) must be similar, and this implicates that segments and are parallel each other. Since is perpendicular to , then so is ; thus, is an altitude of the triangle ABC, and given that H is the intersect of two altitudes ( and ), then H is orthocenter. Thus, H, O and G are collinear.

Important note: The proof above not only showed that H, O and G are collinear, but it also showed an relation between these important triangle’s centers: