The Nine-Point circle for any triangle passes through the three mid-points of the sides, the three feet of the altitudes, and the three mid-points of the segments from the respective vertices to orthocenter. Construct the nine points, locate the center (O) and construct the nine point circle.

*NOTE* The figures on this page seem to look best using Internet Explorer, as opposed to Netscape

    All Constructions will be done using Geometer's Sketchpad 4.0.(GSP) My first step to tackling this problem was to construct a triangle ABC, and find the midpoints of the sides(D, E, F), the feet of the altitudes(H, J, L). I used GSP to find the midpoints of the segments AB, BC, and CA from the construct menu. I then used the same menu to construct the altitudes, the segments perpendicular to sides, and going through the opposite vertices. The Figure 2.1 below is dynamic, and please feel free to grab a vertex of the triangle and play around, noticing that the constructions apply to triangles of all shapes and sizes.
 
 

    The next step is to create the last of the nine points. These points are determined by the altitudes drawn in the previous figure. The pont (N) is called the orthocenter, and it is the point where the three altitudes are concurrent. The final three points of the nine point circle  lie on the midpoints of the segments NA, NB, and NC. I will call these points I, K, M. See Figure 2.2 for this construction. Since i am having fun learning technology, the following figure will also be dynamic in nature.
 
 

      The next step is to find the center of the nine point circle. To find the center of a circle, take two points on the circle, and draw a chord between them. Construct the perpendicular to that chord at it's midpoint (See Figure 2.3a below) If you construct the perpendicular to the chord through the midpoint, the perpendicular will pass through the circle. The tangent line of the circle at this point will be parallel to the original chord. It is a well known fact about circles that a tangent line is perpendicular to the radius at the point of intersection with the circle. This would mean the the perpendicular line is concurrent with the diameter. If this is done with two seperate chords in a circle, the intersection of the two perpendicular lines (or two diameters) will be the center of the circle!!!

    The last step is to draw a nice red circle, the nine-point circle, using (O) as the center. I will use the circle tool, and drag the radius from (O) to (F). (I could have used any of the nine points) I then hid all of the green lines, leaving only the pertinent lines and points. Please observe Figure 2.3 below. Play aroung with the figure and notice that the nine point circle exists for any shape triangle, from right, to acute, to obtuse. The nine point circle will exist for any triangle!!
 
 

    I discovered some things while playing around with the nine-point circle, and Geometer's Sketchpad in general. One of these has to do with equilateral triangles. In this special case, three of the nine points in the nine point circle appear to dissappear. The reason for this occurence is the overall symmetry of this special triangle. In this case (See Figure 2.5) we can see that the centroid, orthocenter, and the circumcenter of triangle ABC are all the same point. This does not occur with any other triangle!! Also, the altitudes, the angle bisectors, and the medians of the angles are all the same lines!!! The nine-point circle (red) also appears to be the same circle as the incircle of the triangle!

      The above page is the nine point circle, as constructed by Jason Blaesing using Geometer's Sketchpad 4.0 on 4/17/2002.