Assignment 4: Centers of a Triangle

Laura Dickerson

Prove that for any triangle, H, G, and C are collinear, and prove that HG=2GC.

To prove that for any triangle, H, G, and C are collinear, let us first look at how they are constructed. The orthocenter, H, is the common intersection of the three lines containing the altitudes, the centroid, G, is the common intersection of the three medians, and the circumcenter, C, is the common intersection of the three perpendicular bisectors.

The diagram to the left shows us a triangle and H, G, and C for that triangle.

H is constructed as the intersection of the altitudes. The altitudes of this triangle are the purple lines.

G is constructed as the intersection of the medians. The medians of this triangle are the red line segments.

C is constructed as the intersection of the perpendicular bisectors. The perpendicular bisectors are the blue lines.

Notice that the line segment HC has been constructed.

Click here for the GSP sketch to see if G is always on this line when any point is moved around.

Why are these points always collinear?

We can show that three points are collinear if we show that the third point lies on the line that runs through the other two points. This is the approach that we will take to prove that the for any triangle, H, G, and C are collinear. We will construct G and C, given a triangle and prove that H lies on the line GC.

The object to the right shows the triangle DEF with the centroid and the circumcenter constructed.

The red line is the line GC.

We next will construct one altitude of triangle DEF. The point of intersection (X) of this altitude and line GC is collinear with the points G and C.

Next, I will show that the point X lies also on the other two altitudes. Then I will show that point X is the orthocenter of the triangle and that the points H, G, and C are collinear.

The next two images show when I construct a perpendicular line from point X to one of the sides of the triangle.

Notice that both these lines also run through the opposite vertex of the triangle

.

Now we can see that the two lines just constructed are altitudes of the triangle DEF, thus the point X is the orthocenter of triangle DEF and should be labeled H

Prove that HG = 2GC.

If we look at the triangle to the right (our original triangle), then we see that it is a mess to work with. If we can show that the triangle with base HG is twice as large as the triangle with CG as the base. So let us just look at these triangles and the lines that form them.

The centroid (G) trisects each median. So in the drawing to the left, the point G trisects the red line segment. We now have a chance to show that these two triangles are similar.

Notice that we are now just looking at the two triangles AHG and ECG.

What do we know?

AH and EC are parallel since they both are segments of lines that are perpendicular to the same line.

AG = 2GE since G trisects AE.

Since angles AGH and EGC are vertical, they are congruent.

Angles HAG and CEG are congruent because they are alternate interior angles of a line intersecting two parallel lines. Likewise, angles AHG and ECG are congruent.

If two triangles have three congruent angles, they are either congruent or similar. Since we know that these two triangles are not congruent, because AG = 2GE, it is similar. We know that corresponding sides of a triangle are also similar with the same ratio. Thus HG = 2GC.