Proof of np center
Nine Point Circle with Euler Line
Use coordinates to prove that the nine-point center is the midpoint of the segment from the orthocenter to the circumcenter.
(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)
Proof:
Create 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 O.
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.
AE is perpendicular to BC, thus
(slope of AE)(slope of BC)=-1.
.
So .
The equation of AE is thus
.
If we plug in the point A=(-a,0) and solve for h, we find that
.
So the orthocenter, H, is located at the point .
Next, we will find the coordinates of the circumcenter, O. To do this, we will find the intersection of two of the perpendicular bisectors . The midpoint of AC occurs at point P. We can easily determine the coordinates of P using the midpoint formula to be . Because P lies on the x-axis, we know that the equation of the line perpendicular to the x-axis, through P, is .
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 .The slope of this line is going to be perpendicular to the slope of BC. So as before, . 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 . By substituting into this equation, we can find the y value of the circumcenter. We can then determine the coordinates of the circumcenter to be .
Finally, we can determine the midpoint of the orthocenter, H, and the circumcenter, O, using the midpoint formula. After some simplification, this becomes:
.
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 . Therefore, the midpoint of the chord FT is . The midpoint of the chord FP is . 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 .
Thus, we see that the nine-point center is the midpoint of the segment from the orthocenter to the circumcenter.