Proving the existence of the Euler line using methods from Coordinate Geometry. I saw a video by Salman Khan, in which he gave a proof of existence of the Euler Line. He proved that the circumcenter, orthocenter and centroid of a triangle are collinear, and used normal geometry to do this. My background is in coordinate geometry, and I want to prove that the Euler line exists, using only methods from coordinate geometry, such as the distance between points formula, the section formulae (internal division), the perpendicular bisector formula, centroid formula and so on. For this, I need to know the coordinates of the circumcenter and the orthocenter, since the coordinates for the centroid are very easy to find and are given by the centroid formula.
So, if I consider a triangle $ABC$ with $A \equiv (x_1, y_1)$, $B \equiv (x_2, y_2)$ and $C \equiv (x_3, y_3)$ coordinates then by using the centroid formula I can find out that the x-coordinate (abscissa) and the y-coordinate (ordinate) of the centroid are following repectively:  $$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$ Let's denote the centroid of the triangle $D$.
Now, using perpendicular bisector formula, with some easy steps I got that the coordinates of the circumcenter of the triangle are the followings: $$\left(\frac{-\frac{1}{2}(x_2^{2}+y_2^{2}-x_3^{2}-y_3^{2})(y_1-y_2)+\frac{1}{2}(x_1^{2}+y_1^{2}-x_2^{2}-y_2^{2})(y_2-y_3)}{(x_1-x_2)(y_2-y_3)-(x_2-x_3)(y_1-y_2)}, \frac{-\frac{1}{2}(x_1^{2}+y_1^{2}-x_2^{2}-y_2^{2})(x_2-x_3)+\frac{1}{2}(x_2^{2}+y_2^{2}-x_3^{2}-y_3^{2})(x_1-x_2)}{(x_1-x_2)(y_2-y_3)-(x_2-x_3)(y_1-y_2)}\right) $$
Let's denote the circumcenter of the triangle $E$.
Then finding the equations of the altitudes of triangle $ABC$, I found out again with some easy steps, that the coordinates of the orthocenter of the triangle are the followings:
$$\left(\frac{(y_2-y_1)\{y_1(y_3-y_2)+x_1(x_3-x_2)\}-(y_3-y_2)\{y_3(y_2-y_1)+x_3(x_2-x_1)\}}{(x_2-x_3)(y_1-y_2)-(x_1-x_2)(y_2-y_3)}, \frac{(x_3-x_2)\{y_3(y_2-y_1)+x_3(x_2-x_1)\}-(x_2-x_1)\{y_1(y_3-y_2)+x_1(x_3-x_2)\}}{(x_2-x_3)(y_1-y_2)-(x_1-x_2)(y_2-y_3)}\right)$$
Let's denote the orthocenter of the triangle $F$.
As you can see, The circumcenter and orthocenter has very difficult coordinates in formation which I tried but can not be simplified any further. As you've already seen, to find from coordinate geometry, I need to solve simultaneous equations to find the intersection of two lines whose equations I got from using the vertices of the triangles, and this leads to a this very complicated Cartesian form of coordinates. As part of my textbook, I need to prove that the orthocenter, centroid and circumcenter is collinear and if I denote those points respectively as $F, D, E$ then prove that the ratio of the line segment $FD$ to $DE$ is $2$ to $1$. That is $$\frac{FD}{DE}=\frac{2}{1}$$
This is basically the gist of the Euler's Line Theorem. I know that the circumcenter of a triangle is the orthocenter of the medial triangle, but this is not helpful for me, maybe you can see why.
One online resource for "What is the formula for circumcentre?" if that helps but sadly no one gives an explicit formula using only the coordinates of the vertices.
Help:
Can someone provide a proof of the fact that the circumcenter, centriod and orthocenter are collinear? And using my coordinates of the Circumcenter, Orthocenter and Centroid using section formula can you prove that $FD : DE = 2:1$ from there?

I have been struggling with this, so I would be grateful if someone
help me work it out.


Links:
https://www.khanacademy.org/math/geometry-home/triangle-properties/triangle-property-review/v/euler-s-line-proof
https://www.quora.com/What-is-the-formula-for-circumcentre
 A: we have to prove that $r_{x}=r_{y}$ where
$\begin{array}{} r_{x}=\frac{x_{F}-x_{D}}{x_{D}-x_{E}} & r_{y}=\frac{y_{F}-y_{D}}{y_{D}-y_{E}}  \end{array}$
coordinates (x,y) of centroid(D), circumcenter(E) and orthocenter(F) as determinants. To get $y_{D}$, $y_{E}$ and $y_{F}$, exchange x for y in the formulas.
$x_{D}=\frac{x_{1}+x_{2}+x_{3}}{3}$
$x_{E}=\frac{1}{2·Det}·\left| \begin{array}{} x_{1}^2+y_{1}^2 & y_{1} & 1 \\ x_{2}^2+y_{2}^2 & y_{2} & 1 \\ x_{3}^2+y_{3}^2 & y_{3} & 1 \\ \end{array} \right|$
$x_{F}=\frac{1}{Det}(\left| \begin{array}{} x_{1} & x_{1}·y_{1} & 1 \\ x_{2} & x_{2}·y_{2} & 1 \\ x_{3} & x_{3}·y_{3} & 1 \\ \end{array} \right|+\left| \begin{array}{} y_{1} & y_{1}^{2} & 1 \\ y_{2} & y_{2}^2 & 1 \\ y_{3} & y_{3}^2 & 1 \\ \end{array} \right|  )$
$Det=\left| \begin{array}{} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \\ \end{array} \right|$
substituting, multiplying by $6.Det$, we get
$r_{x}=\frac{2·[3·(\left| \begin{array}{} x_{1} & x_{1}·y_{1} & 1 \\ x_{2} & x_{2}·y_{2} & 1 \\ x_{3} & x_{3}·y_{3} & 1 \\ \end{array} \right|+\left| \begin{array}{} y_{1} & y_{1}^{2} & 1 \\ y_{B} & y_{2}^2 & 1 \\ y_{3} & y_{3}^2 & 1 \\ \end{array} \right|  )-\left| \begin{array}{} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \\ \end{array} \right|·(x_{1}+x_{2}+x_{3}) ]}{2·\left| \begin{array}{} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \\ \end{array} \right|·(x_{1}+x_{2}+x_{3}) -3·\left| \begin{array}{} x_{1}^2+y_{1}^2 & y_{1} & 1 \\ x_{2}^2+y_{2}^2 & y_{2} & 1 \\ x_{3}^2+y_{3}^2 & y_{3} & 1 \\ \end{array} \right| }$
simplify, thirty-six terms (18 in the numerator and 18 in the denominator): $r_{x}=2$
By analogy: $r_{y}=2$ (as it's just replacing x by y) we get the same result from $r_{x}$ to $r_{y}$.
