The 9-Points Circle (Feuerbach's Circle)

We will start by recalling some high school geometry facts.

1. The line joining the midpoints of two sides of a triangle is parallel to the third side and measures 1/2 the length of the third side of the triangle.

a) Why is the ratio of side AD to side AB 1:2?

b) In the diagram above, DAE is similar to BAC because ....

Since similar triangles have congruent angles, we have that ADEABC.

c) Line DE is parallel to BC because ....

d) Since the ratio of corresponding sides of similar triangles is constant, what is the ratio of side DE to side BC?

2. Four points, forming the vertices of a quadrilateral, lie on a circle if and only if the sum of the opposite angles in the quadrilateral is 180°.

e) An angle inscribed in a circle has measure equal to 1/2 the measure of the arc it subtends. In the diagram above, what arc does the DAB subtend? What arc does the opposite BCD subtend?

f) Since the total number of degrees of the arc of the full circle is 360°, what is the sum of the measures of DAB and BCD?

Three points, not on a line, determine a unique circle. Suppose we have four points, no three on a line. We can pick any three of these four and construct the circle that contains them. The fourth point will either lie inside, on or outside of this circle. Let's say that the three points determining the circle are A, B and C. Call the fourth point D. We have already seen that if D lies on the circle, then m(ADC) + m(ABC) = 180°.

g) What can you say about the size ofADC if D lies inside the circle? What is then true about m(ADC) + m(ABC)?

h) What can you say about the size ofADC if D lies outside the circle? What is then true about m(ADC) + m(ABC)?

A trapezoid is a quadrilateral with two parallel sides. An isosceles trapezoid is one whose non-parallel sides are congruent.

3. The vertices of an isosceles trapezoid all lie on a circle.

Consider the isosceles trapezoid ABCD below, and draw the line through A which is parallel to BC. This line meets CD in a point that we label E.

i) What kind of quadrilateral is ABCE?

j) What does this say about the sides AE and BC?

k) DAE is what kind of triangle?

Because AE is parallel to BC, BCEAED (corresponding angles of parallel lines). And so, by k) this means that BCDADB.

l) Show that DABABC.

This means that the sums of the opposite angles of the isosceles trapezoid are equal.

m) Show that the sums of the opposite angles of an isosceles trapezoid are 180°.

4. In a right triangle, the line joining the right angle to the midpoint of the hypotenuse has length equal to 1/2 the hypotenuse.

Draw the circumscribed circle O about the right triangle ABC with right angle A.

n) What is the measure of the arc subtended by angle A?

o) What kind of line is BC with respect to this circle?

p) Where is the center of the circle?

q) Prove the theorem.


We are now ready to discuss the Feuerbach circle.

For an arbitrary triangle, the 3 midpoints of the sides, the 3 feet of the altitudes and the 3 points which are the midpoints of the segments joining the orthocenter to the vertices of the triangle all lie on a circle, called the nine-points circle.

There is a circle passing through the 3 midpoints of the sides of the triangle, A', B' and C'. We shall show that the other 6 points are on this circle also.

Let D be the foot of the altitude from A. Consider the quadrilateral A'DB'C'. We claim that this is an isoceles trapezoid.

r) Why is A'D parallel to B'C' ?

s) Why is DB' equal in length to 1/2 AC?

t) Why is A'C' equal in length to 1/2 AC?

Since A'DB'C' is an isosceles trapezoid, D must be on the same circle as A', B' and C'. The other altitude feet (E and F) are dealt with similarly.

u) Determine the isosceles trapezoid that contains E and the one that contains F.

Now consider J, the midpoint of the segment joining the orthocenter H to the vertex A. Draw the circle that has A'J as its diameter.

v) Why is A'B' parallel to the side AB?

w) Why is JB' parallel to the altitude CF?

x) Show that A'B' is perpendicular to JB'.

Angle JB'A' is thus a right angle and so, B' must be on the circle with diameter A'J.

y) In an analogous manner, prove that C' is on the circle with diameter A'J.

We therefore have J, A', B' and C' all on the same circle, which is the circle we started with. By repeating this argument starting with the circles having diameters KB' and LC', we can show that K and L are also on this circle.