Trigonometry

Ontario 10 Academic (MPM2D)

Medians and midpoints in a triangle

Lesson

Triangles have several points that could be called *centres. *In this chapter we discuss the *orthocentre*, *incentre*, *circumcentre *and *centroid*.

If we wished to divide a triangle into two regions of equal area, we could do this by drawing a line that passes through a vertex and the midpoint of the opposite side. Such a line is called a *median*.

In the diagram, the red line is a median. The two half-triangles formed by the median have the __ same area__ because they have the

If another median is inserted, it intersects the first. The point of intersection is called the *centroid*. At the centroid, the mass of a triangle made of some uniform solid material would be balanced across lines pointing in two different directions and so the centre of gravity must be at this point.

In the diagram below, a triangle with two medians is drawn with a line passing through the remaining vertex and the centroid. We show, by a similar triangles argument, that this line is also a median. The symbol '$\sim$~' will be used to indicate similarity.

We begin by constructing the line segment $ED$`E``D`, as in the following diagram.

The goal is to show that $BF=FC$`B``F`=`F``C`. We observe first that triangles $AED$`A``E``D` and $ABC$`A``B``C` are similar (Side-Angle-Side) and therefore, $\angle AED=\angle ABC$∠`A``E``D`=∠`A``B``C`, making $ED$`E``D` parallel to $BC$`B``C`.

Since they have the same angles, $\triangle AEG\sim\triangle ABF$△`A``E``G`~△`A``B``F` and since $AB=2AE$`A``B`=2`A``E`, it follows that

$BF=2EG\ \ (1)$`B``F`=2`E``G` (1)

Now, $\triangle ODE\sim\triangle OBC$△`O``D``E`~△`O``B``C` (Angle-Angle-Angle) and since $BC=2ED$`B``C`=2`E``D`, we have $OC=2OE$`O``C`=2`O``E`.

Finally, $\triangle OEG\sim\triangle OCF$△`O``E``G`~△`O``C``F` and since $OC=2OE$`O``C`=2`O``E` we conclude that

$CF=2EG\ \ (2)$`C``F`=2`E``G` (2).

On comparing $(1)$(1) and $(2)$(2), we see that $CF=BF$`C``F`=`B``F`, as required, and therefore, $AF$`A``F` is indeed a median.

We can make a further deduction concerning the centroid. We have already seen that $\triangle OEG\sim\triangle OFC$△`O``E``G`~△`O``F``C` with $FC=2EG$`F``C`=2`E``G`. It follows that $CO=2OE$`C``O`=2`O``E` and the centroid divides the median $EC$`E``C` in the ratio $1:2$1:2. There is nothing special about this particular median: the same reasoning could be applied to each of the others.

So, we conclude that the centroid divides the medians in the ratio $1:2$1:2 in all cases.

A *height *of a triangle is the perpendicular distance from a vertex to the opposite side. The line drawn through a vertex so that it is perpendicular to the side opposite the vertex is called an *altitude*. It extends beyond the triangle in both directions. There are three altitudes (and three *heights*) in any triangle. Any two altitudes must intersect and, in fact, all three altitudes intersect at the same point, which we prove below.

The intersection point of the altitudes is called the *orthocentre*. The orthocentre may be outside of the region enclosed by the triangle. (It is inside if all the angles of the triangle are acute.) In the following diagram, the altitudes are drawn in red.

There are several proofs that the altitudes are concurrent. The following uses similar triangles and begins with the assumption that the three altitudes intersect pair-wise but not at the same point. This assumption leads to a contradiction and so we conclude that it was false and that the altitudes are indeed concurrent.

In the diagram, because of pairs of equal angles, we have $\triangle BCR\sim\triangle BAP$△`B``C``R`~△`B``A``P` and $\triangle BAP\sim\triangle KAR$△`B``A``P`~△`K``A``R`. Hence,

$\triangle BCR\sim\triangle KAR\ \ \left(1\right)$△`B``C``R`~△`K``A``R` (1).

In a similar way, we deduce that

$\triangle BAQ\sim\triangle CHQ\ \ \left(2\right)$△`B``A``Q`~△`C``H``Q` (2).

(Both are similar to $\triangle BHR$△`B``H``R`.)

Using $(1)$(1) and $(2)$(2), we form ratios $\frac{KR}{AR}=\frac{BR}{CR}$`K``R``A``R`=`B``R``C``R` and $\frac{HR}{BR}=\frac{AR}{CR}$`H``R``B``R`=`A``R``C``R`. So, $KR=\frac{BR}{CR}.AR$`K``R`=`B``R``C``R`.`A``R` and $HR=\frac{AR}{CR}.BR$`H``R`=`A``R``C``R`.`B``R`.

The right-hand sides of these equations are identical. So, $KR=HR$`K``R`=`H``R`, a contradiction. Since $K$`K` is on $PA$`P``A` and $H$`H` is on $BQ$`B``Q`, this can only occur if the points $K$`K` and $H$`H` coincide at the intersection of $PA$`P``A` and $BQ$`B``Q`.

When a triangle is inscribed in a circle, the circle is called the *circumcircle* and its centre is called the *circumcentre* of the triangle.

In the diagram below, the perpendicular bisectors of $AB$`A``B` and $AC$`A``C` intersect at $O$`O`. By congruent triangles, it is clear that $OB=OA=OC$`O``B`=`O``A`=`O``C`. So, these three line segments are radii of a circle with centre $O$`O` which passes through the vertices $A$`A`, $B$`B` and $C$`C`.

Since the sides of the triangle are chords of the circle, their perpendicular bisectors pass through the centre of the circle.

Thus, we can state: The perpendicular bisectors of the sides of a triangle are concurrent at the circumcentre.

When a circle is inscribed in a triangle, the circle is called the *incircle*. Its centre is called the *incentre *of the triangle.

The bisectors of angles $A$`A` and $B$`B` of the triangle intersect at $O$`O`. By congruent triangles, we see that the perpendiculars from $O$`O` to the sides of the triangle are equal in length. Therefore, they are radii of a circle with centre $O$`O` that is inscribed in the triangle.

The sides of the triangle are tangent to this circle since they are perpendicular to the respective radii. Therefore, again by congruent triangles, it is clear that line $CO$`C``O` bisects angle $C$`C`. This establishes the theorem: The angle bisectors of a triangle are concurrent at the incentre.

Determine, through investigation, some characteristics and properties of geometric figures