[math][/math](NOTE: Assume all points correspond to Fig. 5.14 on page 131 of the textbook- I tried to rename the points but every time I hover over a point or line it pops up with the name and coordinate and doesn't let me right click)[b][br]Proof:[br][/b]Create [math]\triangle ABC[/math]such that AC lies on the x-axis and B lies on the positive y-axis. That is, A=(-a,0), B=(0,b), and C=(c,0). Consider the nine point circle of the triangle. Let Q be the midpoint of AB, R be the midpoint of BC, and P the midpoint of AC. Let the foot of AB be point D, the foot of BC be point E, and the foot of AC be point F (the origin). Allow H to be the orthocenter and notice that it lies on the y-axis. Let S be the midpoint of HA, T be the midpoint of HB, and U be the midpoint of HC. Furthermore, call the center of the nine point circle N and the circumcenter of [math]\triangle ABC[/math]O.[br][br]To begin,we will find the coordinates of the orthocenter, H. Notice, H lies on the y-axis, thus the coordinates of H are (0,h). Notice, by definition of the orthocenter, the line AE goes through H. So we can determine the value of h by solving for the equation of the line AE. [br]AE is perpendicular to BC, thus[br](slope of AE)(slope of BC)=-1. [br][math]slopeofBC=-\frac{b}{c}[/math]. [br]So [math]slopeofAE=\frac{c}{b}[/math]. [br]The equation of AE is thus [br][math]y=\frac{c}{b}x+h[/math]. [br]If we plug in the point A=(-a,0) and solve for h, we find that[br][math]h=\frac{ac}{b}[/math]. [br]So the orthocenter, H, is located at the point [math]\left(0,\frac{ac}{b}\right)[/math].[br][br]Next, we will find the coordinates of the circumcenter, O. To do this, we will find the intersection of two of the perpendicular bisectors [math]\triangle ABC[/math]. The midpoint of AC occurs at point P. We can easily determine the coordinates of P using the midpoint formula to be [math]P=\left(\frac{c-a}{2},0\right)[/math]. Because P lies on the x-axis, we know that the equation of the line perpendicular to the x-axis, through P, is [math]x=\frac{c-a}{2}[/math]. [br]Next, we will find the equation of the perpendicular bisector of BC. The midpoint of BC is R. Using the midpoint formula we know that [math]R=\left(\frac{c}{2},\frac{b}{2}\right)[/math] .The slope of this line is going to be perpendicular to the slope of BC. So as before, [math]slope=\frac{c}{b}[/math]. By substituting the point R, into the slope-intercept form of the equation of the perpendicular bisector, RO. We can determine that equation to be [math]y=\frac{c}{b}x+\frac{b^2-c^2}{2b}[/math]. By substituting [math]x=\frac{c-a}{2}[/math] into this equation, we can find the y value of the circumcenter. We can then determine the coordinates of the circumcenter to be [math]O=\left(\frac{c-a}{2},\frac{b^2-ac}{2b}\right)[/math]. [br]Finally, we can determine the midpoint of the orthocenter, H, and the circumcenter, O, using the midpoint formula. After some simplification, this becomes: [br][math]midpoint_{OH}=\left(\frac{c-a}{4},\frac{b^2+ac}{4b}\right)[/math]. [br][br]Now, we can determine the center of the nine point circle, N, using the two chords FT and FP. We must first determine the coordinates of T. T is the midpoint of HB, thus [math]T=\left(0,\frac{b^2+ac}{2b}\right)[/math]. Therefore, the midpoint of the chord FT is [math]midpoint_{FT}=\left(0,\frac{b^2+ac}{4b}\right)[/math]. The midpoint of the chord FP is [math]midpoint_{FP}=\left(\frac{c-a}{4},0\right)[/math]. Because FT lies on the y-axis and FP lies on the x-axis, we can easily determine the center of the nine point circle, N, to be [math]N=\left(\frac{c-a}{4},\frac{b^2+ac}{4b}\right)=midpoint_{OH}[/math]. [br]Thus, we see that the nine-point center is the midpoint of the segment from the orthocenter to the circumcenter.