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.