The Nine-Point Circle

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Just for fun, my Geometry class and I decided that we would do a team Blog about this interesting result from Euclidean Geometry: Feuerbach's Nine Point Circle.  We're going to work on this in a leisurely fashion over a long period, so please check in from time to time to see how far we have gone.  We're planning to put in lots of pictures, to make the geometric reasoning clearer, but since it is being written "by committee", it will probably show the usual signs of Committeeishness.

There are a few terms that some readers might not recognize.
+ Vertex: this means the corners.  In the triangle ABC, the vertex A is just the point A, and so on.  The plural is vertices.
+ The foot of a perpendicular.  A perpendicular is from a point to a line.  The foot of a perpendicular is the point where it hits the line to which we're drawing the perpendicular.
+ An altitude is a perpendicular from a vertex to the opposite side.


Introduction
The Nine-Point Circle Theorem is an interesting result in Euclidean Geometry, having to do with a circle that that passes through six important points on any triangle.  Every triangle has several important points associated with it, and usually these points have little to do with each other.  But it just so happens that someone discovered that six of them all lie on a circle.  Furthermore, it turns out that there are three more relatively unimportant points that also lie on this circle.

To explain the points and their significance, we show them in a sequence of diagrams below.

First, we show the midpoints of each of the sides.  We indicate these in RED.

Next, we show the feet of the altitudes from each vertex to the opposite side. We show these in GREEN.

Incidentally, this sketch illustrates that the altitudes meet at one point, which is called the orthocenter, shown as O below.

Finally, we show that the points that lie midway between the orthocenter and the vertices also lie on the circle; we show these in PINK.

And now, the moment you’ve all been waiting for: the actual circle:

We shall actually prove that these points lie on a circle (though it obviously does, according to the picture).

The proof of the existence of the 9-point circle is based on two previous theorems.

The first of these is the Mid-Point Theorem, which says that if XYZ is a triangle, and P is the midpoint of XY, and Q is the midpoint of XZ, then

(i) PQ = 1/2 YZ, and
(ii) PQ is parallel to YZ.

The proof of this is not difficult.

Let XYZ be a triangle, and let P and Q be the midpoints, as described above.  Consider the diagram at right.

To prove this theorem, we need a construction.  Extend PQ to point R, in such a way that PQ and QR are congruent (i.e., equal in length).  Join ZR.  Now triangles PQX and RQZ are congruent by "Side-Angle-Side".

Angles XPQ and ZRQ are congruent by Corresponding Parts.  Using the Alternate Angle Theorem, we know that lines XPY and RZ are parallel.

Consider the second diagram.  As you can see, XP, RZ, and PY are all congruent.  There is a result that says that if PY and RZ are both parallel and congruent, then PR and YZ are also both parallel and congruent.  It also means that the length of PQ is half of the length of YZ!

The second result we need is the interesting fact that if UV is the diameter of a circle, and if it is a side of a triangle whose third vertex, W, is on one of the semicircles, then angle W will be a right angle.  We give a diagram; the result follows from a little angle-chasing (notice that there are two isosceles triangles in the figure).


Now we're ready to prove the theorem.  Unfortunately, the semester is over, the students have dispersed, and I have to write up this part all by myself.


First, let's look at the three altitudes, their feet, and also those points halfway between the orthocenter and the three vertices; i.e, the pink points.  And also the orthocenter itself, which is shown in white.

Look at the triangle OBC. BO is an altitude of the big triangle, ABC, and the pink point is halfway on OB; that is, it's the midpoint of OB.  Similarly the pink point on OC is its midpoint.  So by the Midpoint Theorem, the line joining the two points is parallel to BC.  That's easy.  Also, quite importantly as it happens, BC is perpendicular (90 degrees) to the line AO, so both BC and the line joining the two pink points are perpendicular to AO.  Also, this line joining the pink points is half the length of BC.

Let's try to (temporarily) forget about the pink points, and let's go on to the midpoints of the sides AB, BC and CA, shown in Red below.

Look at the triangle ABC, the original triangle.  Because the red points are midpoints, the line joining the two red points on AB and AC is parallel to BC.  In fact, it is half the length of BC, and is again perpendicular to AO.  So putting the two sketches together,

you see that the two lines we have added to the diagram, both parallel to BC as we showed, are parallel to each other, too.  In addition, they really seem to be opposite sides of a rectangle; we only need to show that the remaining two opposite sides are parallel to AO, which is perpendicular to those two lines shown here.

Well, look at the diagram below.  It emphasizes the triangle BOA.  Because of the Midpoint Theorem, B''N should be parallel to AO.  It is a little peculiar to consider a line that looks "vertical" as the base of the triangle, but of course it's all relative.

The base is AO, which we already know is perpendicular to NM and B''C'', so B''N is perpendicular to NM and B''C''.

Using almost the same argument, we can conclude that C''M is also perpendicular to the two "horizontal" lines.

This is an important discovery.  The fact that NMC''B'' is a rectangle brings us almost halfway to the 9-point circle, because rectangles are concyclic, which means that a circle can be drawn through the four corners of a rectangle.  You draw the two diagonals, and center the circle there.  So here, too, we shall draw a circle through the four points NMC''B''.  What remains is the (non-trivial) task of showing that the other five points also lie on this circle.

We can cheat a little.  Just like the rectangle NMC''B'', we can show that LMA''B'' is also a rectangle!  The arguments are the same:

AB is parallel to A"B", by the Midpoint Theorem applied to OAB, ML is parallel to AB applying Midpoint Theorem to CAB, and A"M is parallel to OC applying the Midpoint Theorem to AOC, and B"L is parallel to OC applying Midpoint to BOC.  So we can draw a circle containing L, M, A'', and B''.

Now here's the punchline: the two circles are the same.  The reason is that both their centers are the midpoint of B"M.  The two circles have the same center, and both pass through M, therefore they're the same circle!  Here's a pictorial summary of the facts we have revealed thus far:

We can count 6 points on this circle: L, M, N, the midpoints of the three sides of the big triangle; A", B", C", the three midpoints of the segments AO, BO and CO.  Where are the other three points on the circle?  These are the feet of the three perpendiculars from A, B, C, respectively, to the three opposite sides.  Let's call them A', B', C' respectively.  Here is a diagram showing just the circle, the three newest points, and a couple of other points that we need:

Remember A"L is one of the diameters of the circle.  It was a diagonal of the second rectangle.  Now we use the other theorem that we proved, about a triangle inscribed in a semicircle.  But we use it in reverse; in other words, we use what is called its converse.  This says that if a right triangle has the diameter of a circle as its hypotenuse, then the vertex with the right angle must lie on the circle.  In the present case, we know that A"A'L is a right angle, because AOA' is an altitude, and so perpendicular to BC.  This makes A"A'L a right triangle, whose hypotenuse, A"L is a diameter of our circle.  So A', which is where the right angle is, must lie on the circle!

Similarly, using the right triangle C"NC', whose hypotenuse is C"N, we conclude that C' must lie on the circle.  Finally, using the right triangle B'B"M, whose hypotenuse is B"M, we conclude that B' is also on the circle.  So we have all nine points on the circle, which proves the theorem.

Gosh, that was a lot longer than I expected, especially because of the trillion diagrams that make it a little easier to follow; and I daresay my readers would have preferred that we had included a lot more diagrams.