The Nine Point Circle

The nine point circle (also known as Feuerbach's circle) is one of the most interesting topics in elementary geometry. It doesn't require much prerequisites and so this post can be read by a high school student who has some idea about geometry of circle and triangle.

Just a reminder: All three altitudes of a triangle meet at a single point, which we call orthocenter and a set of points is said to be concyclic if they all lie in a circle.

Consider $\triangle ABC$ and let $O$ be the orthocenter (shown in green). Let $A_a$ be the feet of altitude from vertex $A$ on side $BC$, $A_m$ be the midpoint of side $BC$ and $A_o$ be the midpoint of line segment $AO$ (we have used the colors red, blue and orange respectively for these points). Similar notations corresponding to vertex $B$ and $C$. Then the nine points (which sometimes may not be distinct)  $A_o, B_o, C_o$ (midpoints between orthocenter and vertices) $A_m, B_m, C_m$ (midpoints of sides) and $A_a, B_a, C_a$ (feet of altitudes) lie in a same circle which is commonly called the nine point circle (shown with dotted circle).

Try interacting with the GeoGebra applet given below ($\Omega$ is the center of nine point circle):

To prove this we need some elementary results:

• A convex quadrilateral is cyclic (ie. its vertices are concyclic) if and only if its opposite angles are supplementary (ie. $180^{\circ}$). One half of this result is Euclid's proposition $22$ in Book III.
• Line segment formed by connecting the midpoints of two sides of a triangle will be parallel to the third side and have half of its length. This called midpoint theorem.
• The center of the circumcircle of a right triangle lies on its hypotenuse. This is converse of a famous result called Thales' theorem.

Let us start by constructing line segments $C_mB_m$, $B_oC_o$, $C_mB_o$ and $C_oB_m$.

Now in $\triangle OBC$, we use midpoint theorem to conclude $B_oC_o$ is parallel to $BC$. Similarly in $\triangle ABC$ we get $C_mB_m$ is parallel to $BC$. Similarly by considering $\triangle AOC$ and $\triangle AOB$, we get $C_mB_o$ and $C_oB_m$ are parallel (since both are parallel to $AO$ by midpoint theorem). Since line $AO$ (ie line $AA_a$ is perpendicular to $BC$, we get $C_oB_m$ is perpendicular to $C_mB_m$. So, $\square C_mB_mC_oB_o$ is a rectangle and in particular is a cyclic quadrilateral. Moreover, by converse of Thales' theorem, $C_mC_o$ and $B_mB_o$ are the diameters of the circle passing through vertices of $\square C_mB_mC_oB_o$. Again by converse of Thales' theorem, $C_a$ and $B_a$ also lie on this circle.

We have proved that $C_m,B_m,C_o,B_o,C_a, B_a$ lie on a circle with diameters $C_mC_o$ and $B_mB_o$. Similarly, we can show $A_m,B_m,A_o,B_o,A_a, B_a$ lie on a circle with diameters $A_mA_o$ and $B_mB_o$ and $A_m,C_m,A_o,C_o,A_a, C_a$ lie on a circle with diameters $A_mA_o$ and $C_mC_o$. Looking at the diameters of these three circles, we see that any two of them has a common diameter. So these three circles must coincide and is indeed the required nine point circle.

References:

•  A Sequel to the First Six Books of Euclid by John Casey
• David Joyce's web version of Euclid's Elements