1. SPECIAL TRIANGLE POINTS

We begin by looking at three special points which are associated with any triangle in the plane, and an interesting property they share. These points are the circumcentre, orthocentre, and centroid.

Circumcentre

Let ABC be any triangle, and let A', B', C' be the midpoints of the sides BC, CA, AB respectively. We claim that:

The perpendicular bisectors of the sides at A', B', C' meet in a point O called the circumcentre.

Can you prove this?

Referring to the diagram, use two of the perpendicular bisectors to determine O, then drop a perpendicular from O to the third side of the triangle. What needs to be proved?

Perhaps you need a further hint?

                                        
The perpendicular bisector of a chord AB is the locus of points which are equidistant from A and B. It is a straight line which is perpendicular to AB, and passes through the midpoint of AB.

































We have asked to show that:

The perpendicular bisectors of the sides at A', B', C' meet in the point O.

In order to prove this we take the perpendicular bisectors C'O and B'O of sides AB and AC to meet in O, and drop the perpendicular OA' to side BC.

Our result will be proved if we can show that A' is the midpoint of side BC.

Proof OB = OA (since OC' perpendicularly bisects AB)
                   = OC (since OB' perpendicularly bisects AC)

Now OB = OC implies that O lies on the perpendicular bisector of BC. Hence OA' is the perpendicular bisector of BC and A' is the midpoint of BC as required.

This completes the proof.

We observe that since OA = OB = OC, O is the centre of the (unique) circumcircle which passes through the vertices of the triangle. O is thus called the circumcentre of the triangle.

                                         









































Orthocentre

Let ABC be any triangle, and let the altitudes of the triangle be AD, BE, CF respectively. We claim that:

The altitudes of triangle ABC meet in a point H called the orthocentre.

Can you prove this?

There are various ways of establishing this result, none of them very obvious. However, there is a very clever simple proof, which comes with a hint! We notice that the altitudes are line segments which are perpendicular to the sides of the triangle. In our work on circumcentre, we found a result about lines which had this property.

So, could we create a new triangle, somehow related to the given ABC, for which the given altitudes are in fact perpendicular bisectors? This is not hard, but might require a little insight. And having found this triangle, can we use it to obtain our required result?

When you think you have the answer, check it out here.

                                             
An altitude of a triangle is the line segment drawn from any vertex to meet the opposite side of the triangle at right angles. The word is sometimes also used to denote the length of such a segment, particularly when the relevant ‘opposite side’ is described as the base of the triangle.































We have asked to show that

The altitudes of triangle ABC meet in a point H called the orthocentre.

As in the diagram, construct A"B"C" with sides parallel to the sides of ABC, and passing through vertices A, B and C.

Now since the opposite sides of a parallelogram have equal length, C"A = BC = AB", so A is the midpoint of B"C". Thus AH is the perpendicular bisector of side B"C" of A"B"C". We can argue similarly for the other sides, and from our previous work, these perpendicular bisectors all meet at point H, the circumcentre of A"B"C".

Hence the altitudes AD, BE, CF of ABC all meet in point H. This point is called the orthocentre of ABC.

What a clever and easy way of obtaining this result!

DEF is sometimes called the pedal triangle of ABC (its vertices are the feet of the altitudes).

                                             
The prefix ortho is used mathematically to mean ‘right’ as in ‘orthogonal’ for two lines meeting at a right angle. Curiously, its normal meaning is ‘straight’ as in ‘orthodox’ (= ‘straight’ doctrine) or ‘orthoptics’ (= the study of straightening eyes).





























Centroid

Let ABC be any triangle, and A', B' C' the midpoints of sides BC, CA, and AB respectively, as in the diagram.

The segments AA', BB', CC' are called the medians of the triangle.

We claim that:

The medians of triangle ABC meet in a point G called the centroid.

See if you can prove this. It is not hard, and there are various possible methods. You might try showing that G divides a median in the ratio 2:1.

Click here for a helpful hint ...

                                         







































We have claimed that

The medians of triangle ABC meet in a point G called the centroid.

Hint: the diagram gives a good start to one method of establishing this result. Can you show that BG = 2GB' ?

Proof   Start with ABC and medians BB', CC'. We shall need to use some well known angle properties of parallel lines,
and also the fact that C'B' is parallel to BC and half its length. Because BC // C'B', the marked alternate angles are equal. Hence BCG and ABC are similar – they have the same shape, and the lower triangle is in fact twice the size of the upper triangle. It follows that BG = 2GB'. In other words, median CC' meets median BB' at a point G two-thirds of the way along its length. Since we could apply the same argument to medians BB' and AA', we conclude that all the medians meet at the point G – the centroid of ABC. This is the required result.

                                      
Suppose we have two parallel lines and a common transversal (crossing line).

When lines l and m are parallel, a and b are called corresponding angles, b and c are alternate angles. It is an axiomatic property of parallel lines that a = b, and b = c. In any case, a and c are called vertically opposite angles, and a = c.
Theorem: The segment joining the midpoints of two sides of a triangle is parallel to the base and equal to half its length.

Proof Since we know more about parallel lines, we try working backwards.

Let ABC be a given triangle, R the midpoint of AB, and
RQ // BC, RP // AC as in the diagram. By properties of parallel lines, the marked angles are equal, so ARQ is congruent to RBP (A.S.A. test).

Hence AQ = RP = QC. So a line through R parallel to the base passes through midpoint Q. Since midpoint Q has a unique position, the first part of the theorem is proved.

Also, RQ = BP = PC, so BC = 2 RQ as required.




























Euler Line

Our diagram is starting to look quite ridiculous!

For ABC we have now defined the circumcentre O, the orthocentre H and the centroid G. We claim that

The three points O, H, G all lie on a straight line, the Euler line, and furthermore, G divides HO in the ratio 2:1.

From our work thus far, ABC is mapped to A"B"C" by an enlargement with centre G. Under this mapping, the orthocentre H of
ABC will map to the orthocentre of
A"B"C". As we have seen, this is the circumcentre O of ABC. Hence points H, G and O are collinear.

The ratio 2:1 follows from the fact that A"B"C" is twice the size of ABC.

                                                     
































Extensions

The best way to come to grips with this sort of geometry is just to play with it. Here are some simple exercises. See if you can provide the answers.

1. Show AH = 2A'O.

2. G is the centroid of A'B'C'.

3. A is the orthocentre of HBC. (When does the orthocentre lie outside a triangle? on a triangle?)

4. Is it possible for any two of the three ‘triangle points’ to coincide? What happens to the third point in this case?

5. Show that GH = 2 GO by considering HGA and OGA'.

6. The triangle A"B"C" is constructed as before by drawing parallel lines to the sides of ABC through the opposite vertices. What is the circumcentre of this larger triangle?
                                                                                                                 Hints and Solutions ...

For looking up ...

http://en.wikipedia.org/wiki/Euler's_line

http://aleph0.clarku.edu/~djoyce/java/Geometry/eulerline.html

                                          










































Hints and Solutions

1. Show AH = 2A'O.

• AHG is similar to A'OG and twice its size
    (AG = 2GA').

2. G is the centroid of A'B'C'.

• Observe that the medians of A'B'C' lie along
   the medians of ABC.

3. A is the orthocentre of HBC. (When does the orthocentre lie outside a triangle? on a triangle?)

• H satisfies the definition of orthocentre for HBC.
   Think about an obtuse-angled triangle; a right triangle.

4. Is it possible for any two of the three ‘triangle points’ to coincide? What happens to the third point in this case?

• Two of the ‘triangle points’ coincide when the triangle is equilateral, and then all three points coincide.

5. Show that GH = 2 GO by considering HGA and OGA'.

• Argue as in Q1.

6. The triangle A"B"C" is constructed as before by drawing parallel lines to the sides of ABC through the opposite vertices. What is the circumcentre of this larger triangle?

• The circumcentre is H. Just check that A, B, C are midpoints of the sides of the new traingle.