Use coordinates to prove that the nine-point center is the midpoint of the segment from the orthocenter to the circumcenter. This shows that the nine-point center lies on the Euler line.[br][br][br][u][b]Proof:[/b][/u] Let ABC be a triangle for which A = (a, 0), B = (0, b), and C = (c, 0). Previously, we found the orthocenter, D, to be located at the following coordinate: D = (0, [math]-\frac{ac}{b}[/math]). Similarly, we found the center of the nine-point circle, E, to be located at the following coordinate: E = ([math]\frac{\left(a+c\right)}{2}[/math], [math]\frac{\left(b^2-ac\right)}{4b}[/math]) -- [in the book].[br][br][br]First, let us determine the coordinates of the circumcenter through a series of steps:[br][br][u][b]Step 1:[/b] Determine the midpoints of each side of ABC using the midpoint formula (denoted by black points) [/u][math]\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)[/math][u]:[/u][br]Midpoint of AB: [math]\left(\frac{a+0}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)[/math].[br]Midpoint of AC: [math]\left(\frac{a+c}{2},\frac{0+0}{2}\right)=\left(\frac{a+c}{2},0\right)[/math].[br]Midpoint of BC: [math]\left(\frac{0+c}{2},\frac{b+0}{2}\right)=\left(\frac{c}{2},\frac{b}{2}\right)[/math].[br][br][br][u][b]Step 2:[/b] Determine the slope of each side of triangle ABC using [/u][math]m=\frac{y_1-y_2}{x_1-x_2}[/math][u]:[/u][br]Slope of AB: [math]\frac{0-b}{a-0}=\frac{-b}{a}[/math].[br]Slope of AC: [math]\frac{0}{a-c}=0[/math].[br]Slope of BC: [math]\frac{b-0}{0-c}=\frac{-b}{c}[/math].[br][br][br][u][b]Step 3:[/b] Determine the slope of a line perpendicular to each side of the triangle using[/u][u]:[br]Note: take the reciprocal of each slope in Step 2 and multiply by -1:[br][/u][br]Slope of line perpendicular to AB: [math]\frac{a}{b}[/math].[br]Slope of line perpendicular to AC: undefined.[br]Slope of line perpendicular to BC: [math]\frac{c}{b}[/math].[br][br][u][b]Step 4:[/b] Determine the equation of the lines of the perpendicular bisectors of each side of ABC:[br]Note:[/u] In order to do this, we will use information from Step 1 and Step 3 in coordination with the point-slope formula [math](y-y_1)=m(x-x_1)[/math]. [br][br]Perpendicular line to AB: [math]\left(y-\frac{b}{2}\right)=\frac{a}{b}\left(x-\frac{a}{2}\right)\Longrightarrow y=\frac{a}{b}x+\frac{-a^2+b^2}{2b}[/math].[br]Perpendicular line to AC: [math]x=\frac{a+c}{2}[/math].[br]Perpendicular line to BC: [math]\left(y-\frac{b}{2}\right)=\frac{c}{b}\left(x-\frac{c}{2}\right)\Longrightarrow y=\frac{c}{b}x+\frac{-c^2+b^2}{2b}[/math].[br][br][u][b]Step 5:[/b] Now, using the equations discovered in Step 4, we can use them to find the circumcenter, T, of ABC. Using the equations, let us find their point of intersection by setting two of them equal to one another:[/u][br][br][math]\frac{a}{b}x+\frac{-a^2+b^2}{2b}=\frac{c}{b}x+\frac{-c^2+b^2}{2b}[/math][br][math]\Longrightarrow\frac{a}{b}x-\frac{c}{b}x=\frac{-c^2+b^2}{2b}-\left(\frac{-a^2+b^2}{2b}\right)[/math][br][math]\Longrightarrow\frac{a-c}{b}x=\frac{-c^2+b^2-b^2+a^2}{2b}=\frac{a^2-c^2}{2b}[/math][br][math]\Longrightarrow x=\frac{a^2-c^2}{2b}\cdot\frac{b}{a-c}=\frac{\left(a+c\right)\left(a-c\right)}{2}\cdot\frac{1}{\left(a-c\right)}=\frac{a+c}{2}[/math].[br][br]Now, we know the x-coordinate of the point of intersection.[br][br][u][b]Step 6:[/b] Determine the y-coordiante of the point of intersection by substituting the x-value found in Step 5 into one of the equations from Step 4:[br][/u][br][math]y=\frac{c}{b}\left(\frac{a+c}{2}\right)+\frac{-c^2+b^2}{2b}=\frac{ac+c^2}{2b}+\frac{-c^2+b^2}{2b}=\frac{ac+c^2-c^2+b^2}{2b}=\frac{ac+b^2}{2b}[/math].[br][br]Therefore, we can conclude that the circumcenter, T, of ABC is located at T= [math]\left(\frac{a+c}{2},\frac{ac+b^2}{2b}\right)[/math].[br][br][u][b]Step 7:[/b] Determine the midpoint of the circumcenter and the orthocenter:[/u][br]Recall that the orthocenter, D, is located at D = (0, [math]-\frac{ac}{b}[/math]), and the circumcenter is located at [br]T= [math]\left(\frac{a+c}{2},\frac{ac+b^2}{2b}\right)[/math].[br][br]Using the midpoint formula, we see that the midpoint of DT is as follows:[br][math]\left(\frac{\frac{a+c}{2}+0}{2},\frac{\frac{ac+b^2}{2b}-\frac{ac}{b}}{2}\right)=\left(\frac{a+c}{4},\frac{ac+b^2-2ac}{2b}\cdot\frac{1}{2}\right)=\left(\frac{a+c}{4},\frac{b^2-ac}{4b}\right)[/math]. [br][br]Note that from the given information, we know that then nine-point center, E, is also located at [br]E = ([math]\frac{\left(a+c\right)}{2}[/math], [math]\frac{\left(b^2-ac\right)}{4b}[/math]).[br][br]In conclusion, it follows from the argument above that the nine-point center lies on the Euler line.[br]