Assignment 4: Centers of a Triangle

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**

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**.

**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**.

**AGH** and **EGC** are vertical, they are congruent.

**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**.

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