The Nine-Point Circle

By Pei-Chun Shih

In
this write-up, I am going to explore some geometric features about the
nine-point circle. It gets the name, the nine-point circle, because it passes
through nine significant points of any triangle: the three midpoints of the
sides, the three points where the altitudes meet with the sides, and the three
midpoints of the segments from each vertex to the orthocenter.

Construct
a nine-point circle with the center N:

First,
in an acute △ABC, I constructed a medial triangle DEF formed by the
three midpoints of △ABCÕs three
sides. Second, in the same △ABC,
I constructed an orthic triangle GHI formed by the three feet of the altitudes
of △ABC. Last, I constructed the third △KML inside the original △ABC by connecting the three midpoints of the segments
from each vertex of △ABC to its
orthocenter. Then I found the circumcenter and constructed the circumcircle for
each of the three triangles mentioned above. Surprisingly, these three
secondary triangles have the same circumcenter (denoted as N) and therefore the
same circumcircle.

The
nine points D, E, F, G, H, I, K, L, and M associated with △ABC lie on one circle. The circle consists these nine
points is called the nine-point circle.

Construct
the centroid (G), orthocenter (H), circumcenter (C), and incenter(I) of the triangle
ABC and observe the relationships of these points with N for different shaped
triangles:

1.
When △ABC is an acute triangle, the four points G, H, C and
N lie on the same line.

2.
When △ABC becomes a equilateral triangle, these points, G,
H, C, I and N, overlap. Besides, the medial triangle DEF and the orthic
triangle GHI are congruent.

3.
When △ABC becomes a right triangle, the points C, G, N, and
H lie on the same line. Besides, point C lies on the midpoint of the hypotenuse
of △ABC and point H lies on the vertex of the right angle.

4.
When △ABC becomes an isosceles triangle, all the five points
C, G, I, N, and H lie on the altitude from the vertex angle to its base of △ABC.

5.
When △ABC becomes an obtuse triangle, the points C, G, N,
and H lie on the same line with points H and C outside △ABC.

Prove
that the altitudes of a triangle are concurrent.:

Let
CF, AD, and BE be the altitudes of a triangle ABC.

Let
the point H be the intersection of AD and BE.

Let
the point F' be the intersection of the ray CH and.

Since
AD⊥BC and BE⊥AC, then∠ADB = 90 degree
=∠AEB.

Consider △ABE and △ABD.

△ABE and
△ABD are
right triangles and share the hypotenuse.

Therefore,
we can viewas the diameter of the circle c1 which has concyclic points
A, B, D, and E.

So∠ADE =arc AE =∠ABE by the arc
angle theorem.

Also, ∠HEC = 90degree =∠HDC.

So,
H, D, C, and E are concyclic points on circle c2 for the same reason as above.

Therefore, ∠ADE =∠HDE
=arc HE =∠HCE
=∠ABE.

Consider △ABE and △ACF'.

△ABE and
△ACF' are
similar triangles by the Angle-Angle similarity theorem since ∠ABE = ∠ACF' and ∠BAE =∠CAF'.

Therefore, ∠AEB = 90 degree = ∠AF'C.

However,
we have been given that CF is perpendicular to AB.

So ∠AFC = 90 degree and ∠F'CF = 0 degree.

Hence,
F and F' are the same points and CFÕ = CF.

Since
CFÕ is the extension of CH and CFÕ = CF, CF must pass through the point H where
AD and BE intersect each other.

Thus,
the altitudes of a triangle are concurrent.