Exploring Centers of Triangles

Within The Medial Triangle

Centers of a Triangle

Unlike a circle, which has only one defined center, in any given triangle, there are several centers. The centers are formed by the intersections of various lines within the triangle.


The centroid (G) of a triangle is formed by the intersection of the three medians of the triangle.

The orthocenter (H), is formed by the intersection of the three altitudes of the triangle.

The circumcenter (C), is formed by the intersections of the perpendicular bisector of each side of the triangle.

The incenter (I), is formed from the intersection of the angle bisectors of each angle of the triangle.


Each of these centers can be found within any given triangle. When each center is found for the same triangle, we can begin to see relationships between these four points.


Notice first of all that points H, G, and C are collinear. This will always be true for any size triangle. Furthermore, G will always be between points H and C. As shown by the calculation above, G divides segment HC so that the ratio of HG : GC = 2:1. This relationship will also hold true for all triangles. The line that contains this segment is known as the Euler line. Choose this link to see the proof for the ratio of the Euler line.

You may notice that the point I is non collinear to the other points in this particular case. However, if the triangle is isoceles, all four points will be collinear. Furthermore, if the triangle is equilateral, all four points will be concurrent. Click on the picture below to open a GSP file showing an animation of this idea.

The Medial Triangle

Within any given triangle, one can construct the medial triangle by connecting the three midpoints of the original sides of the triangle.

We can now locate each of the four centers of this triangle as well. In finding C, G, H, and I for both the original triangle and its medial triangle, we see this:

There are some interesting things happening in this diagram that need to be discussed. The first and most interesting fact is that some of the four points in each triangle are the same. The second idea has to do with the ratios of the segments formed by points I G and I'.

The circumcenter of the original triangle (C) is the same point as the orthocenter of the medial triangle (H'). Similarly, the centroid of both the original triangle (G), as well as the medial triangle (G') are the same point.

It makes sense that the centroids are both the same. The line that contains Z', the midpoint of side XY, will also contain Z'',the midpoint of segment X'Y'. Likewise the line that contains Y' will also contain Y'', the midpoint of segment X'Z', and the line that contains X' will also contain X'', the midpoint of segment Y'Z'. Since the three lines containing X', Y', and Z', are the same lines as the ones containing X'', Y'', and Z'', and those original three lines intersect at G, it follows that the lines containing the other midpoints should do so as well.

Now to discuss the reason for the coincidence of H' and C. We know that the sides of the medial triangle are parallel to the corresponding sides. That is that segment X'Z' is parallel to segment ZX, and so on. Since this is true, any line perpendicular to segment ZX will also be perpendicular to X'Z'. Point C is the intersection of lines Y'C and X'C. Line Y'C is perpendicular to XZ, so it is also perpendicular to Z'X'. since this line is drawn through Y', this line is not only the perpendicular bisector of XZ, it is also the altitude drawn to Z'X'. Following this thought for each other perpendicular line, we find that the three lines that intersect to form C, are the same lines drawn to form H'.

The next thing to look into is the ratio of segments within our original diagram.

As you can see from the calculations that have been done, three different pairs of segments are formed in the ratio of 2:1. The fact that HG : GC is 2:1, and that H'G' : G'C' is also 2:1 are not surprising as they are segments that are a part of the Euler line for each of the two triangles. However, the fact that the segments formed by the two incenters(I and I') and the centriod (G/G') are also in the ratio of 2:1. This can be explained very much in the same manner as we proved the ratios of the Euler line. With the constructions shown below, we can use a similar proof to the ones above to show that triangles AGI' and C'GI are similar by AA similarity, and since AI' : IC' = 2 : 1, then I'G : GI is also 2:1.





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