Geometry

Hong Kong

Stage 1 - Stage 3

Lesson

There are four main centres of a triangle:

The centroid is the point of intersection of the three medians. A median is a line drawn from a vertex to the midpoint of the side opposite that vertex.

The orthocentre is the point of intersection of the three altitudes. An altitude is a line drawn from a vertex that meets the side opposite that vertex at right angles.

The circumcentre is the point of intersection of the perpendicular line bisectors. A perpendicular line bisector is the line drawn from the centre of the side of the triangle and perpendicular to it. The orthocentre is the centre of a circumscribing circle shown in blue.

The incentre is the point of intersection of the vertex angle bisectors. A vertex bisector is a line drawn from a vertex drawn so that it bisects the internal angle at that vertex. The incentre is the centre of the inscribed circle. This circle is tangent to the three sides of the triangle, as shown in the diagram.

Note that the angle bisectors don't necessarily pass through the points of tangency of the circle with the triangle. That will only happen when the triangle is equilateral.

Some students are required to know the proofs of some of these, check with your course or teacher, or maybe you are just interested! If so, then go here to read them.

There are a number of proofs that demonstrate that the three medians always intersect at a single point. This point, called the centroid, is also referred to as the centre of gravity due to the fact that it is the point of balance for the triangle.

If we label the vertices of the triangle as $A,B$`A`,`B` and $C$`C` and attach coordinates to each of these, so that the vertices are addressed as $A\left(x_1,y_1\right),B\left(x_2,y_2\right)$`A`(`x`1,`y`1),`B`(`x`2,`y`2) and $C\left(x_3,y_3\right)$`C`(`x`3,`y`3), then the centroid is located at $G\left[\frac{1}{3}\left(x_1+x_2+x_3\right),\frac{1}{3}\left(y_1+y_2+y_3\right)\right]$`G`[13(`x`1+`x`2+`x`3),13(`y`1+`y`2+`y`3)].

In other words, the average of the $x$`x` and $y$`y` parts of the coordinates of the vertices becomes the coordinates of the centroid.

We can develop a proof of the concurrency of the medians in the following way.

Label the points $A,B,C,D,E,F$`A`,`B`,`C`,`D`,`E`,`F` as shown in the figure, and draw in two medians $AD$`A``D` and $BE$`B``E`, and the extra line $ED$`E``D`.

First observe that $AD$`A``D` and $BE$`B``E` must intersect each other *inside the triangle* at a point, say $G$`G`.

Since $|CA|:|CE|$|`C``A`|:|`C``E`| and $|CB|:|CD|$|`C``B`|:|`C``D`| are both in the ratio $2:1$2:1, then because angle $C$`C` is common to triangles $CED$`C``E``D` and $CAB$`C``A``B`, by similar triangles we also know that $|AB|:|ED|$|`A``B`|:|`E``D`| is also $2:1$2:1, with $AB$`A``B` parallel to $ED$`E``D`.

This implies that, in respect of the two transversals $AD$`A``D` and $BE$`B``E`, the ratios $|AG|:|GD|$|`A``G`|:|`G``D`| and $|BG|:|GE|$|`B``G`|:|`G``E`| are also $2:1$2:1. Note that this is true because the two triangles $ABG$`A``B``G` and $DEG$`D``E``G` are similar.

The same observation can be made with a different pair of medians, say $|CF|$|`C``F`| and $|BE|$|`B``E`|, and their point of intersection must be $G$`G` again. since there is only one point which divides $BE$`B``E` in the ratio $2:1$2:1.

This completes the proof.

Before we look at the proof of the concurrency of the bisectors, we need to realize that, unlike the centroid, the circumcentre can actually be positioned on or outside of the triangle.

In the case of an obtuse triangle, the circumcentre lies outside of the triangle as shown here:

In the case of a right-angled triangle, the circumcentre lies on the hypotenuse as shown here:

Irrespective of the type of triangle, and the location of the circumcentre, the proof is essentially the same.

The proof hinges on a particular observation, which we need to deal with first.

Consider this diagram, which shows a perpendicular bisector of the line segment $AB$`A``B`.

Any point (two examples $P$`P` and $Q$`Q` are shown) on this perpendicular bisector will be equidistant from the endpoints $A$`A` and $B$`B`. Conversely, if a point is equidistant from $A$`A` and $B$`B`, then it must be located on the perpendicular bisector (extended if necessary).

This simple but crucial observation can be verified either by a congruency argument, or simply because the perpendicular bisector acts as a line of symmetry.

Now for the proof of concurrency:

Suppose for any triangle $ABC$`A``B``C` (see below), on *any two* of its sides, say $AC$`A``C` and $BC$`B``C`, perpendicular bisectors are drawn to intersect at the point $O$`O`.

Then, $|AO|=|CO|=x$|`A``O`|=|`C``O`|=`x` (shown in green), and since $|CO|=|BO|$|`C``O`|=|`B``O`|, it must be true that $|BO|=x$|`B``O`|=`x` (shown in blue).

Thus we have established that $|AO|=x$|`A``O`|=`x` (green) and $|BO|=x$|`B``O`|=`x` (blue), and hence a third perpendicular bisector drawn from $AB$`A``B` must contain the point $O$`O`.

Therefore all three perpendicular bisectors pass through $O$`O` establishing their concurrency.